Answer:
d = 9 [m]
Explanation:
This is a problem where we must be clear about the concept of distance, which tells us that it is a measure of the space traveled on a trajectory.
Therefore:
d = 2 + 7
d = 9 [m]
The directions are not taken into account, as it is the sum of the displacements traveled, regardless of their directions.
A book is sitting on a table. Which of the following is true about the table?
Answer:
Its pulling down on the book
Explanation: if it was pushing up the book would be floating and the other choices don't make sense
Help me please..
When the slope of a velocity vs. time graph is negative and constant,
what type of motion is occurring?
A) No motion
B) Constant speed
C) Acceleration
D) Constant Velocity
E) Going in circles
A school bus has a mass of 18,200 kg. The bus moves at 13.5 m/s. How fast must a 0.142-kg baseball move in order to have the same momentum as the bus?
Answer:
bus momentum
p_bus= m_bus x v_bus
=18,200 x 16.5
basball momentum
pball=mball x vball
=0.142 x v
p_bus = pball
18200 x 16.5 = 0.142 x v
v=(18200 x 16.5)/0.142
v is the answer for baseball
Explanation:
⚠️not my answer tryna be honest here⚠️
The momentum of the bus of 18200 kg and velocity of 13.5 m/s is 245700 Kg m/s. To have equal momentum the base ball with 145 g have to throw in a speed of 1.7 × 10 ⁶ m/s.
What is momentum?Momentum of a moving body is the product of mass and velocity. Thus it have the unit of g m/s or Kg m/s. Momentum is a vector quantity and thus having magnitude and direction.
Given that one bus is having a mass of 18200 Kg and 13.2 m/s speed. The momentum is:
p = mv
=18200 kg × 13.5 m/s
= 245700 Kg m/s
To have a momentum of 245700 Kg m/s the base ball with 0. 142 g have to have a velocity = 245700 Kg m/s / 0.142 g
=1.7 × 10 ⁶ m/s
Hence, the baseball weighs0. 142 g have to move in 1.7 × 10 ⁶ m/s
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A 5 kg block rests on an inclined plane with a coefficient of static friction equal to 0.30. What is the minimum angle at which the block will begin to slide
Answer:
[tex]\theta = 16.70 ^{\circ}[/tex]
Explanation:
The coefficient of static friction is equal to the tangent of the minimum angle at which an object will begin to start sliding down a ramp.
[tex]\displaystyle u_s=\frac{F_f}{F_N} = \frac{F_g\ \text{sin}\theta}{F_g\ \text{cos} \theta} = \text{tan} \theta[/tex]Since we are given the coefficient of static friction we can solve for the minimum angle that the block will begin to slide.
Let's solve for the force of gravity that is acting on the block. The force of gravity is also known as the weight force, which can be calculated by using w = mg.
[tex]w=mg[/tex]We are given the mass of the block (kg) and we know that g = 9.8 m/s².
[tex]w=(5)(9.8) = 49 \ \text{N}[/tex]Now we can use this force in the equation:
[tex]\displaystyle u_s = \frac{F_g \ \text{sin} \theta }{F_g \ \text{cos} \theta}[/tex]Plug [tex]\displaystyle u_s = 0.30[/tex] and 49 N into the equation.
[tex]\displaystyle 0.30 = \frac{(49) \ \text{sin} \theta }{(49) \ \text{cos} \theta}[/tex] [tex]0.30=\text{tan} \theta[/tex]Notice that the gravitational force cancels out in the end, so we can actually start with [tex]0.30=\text{tan} \theta[/tex].
Evaluate this equation by taking the inverse tangent of both sides of the equation.
[tex]\text{tan}^-^1 (0.30) = \text{tan}^-^1 (\text{tan}\theta)[/tex] [tex]\text{tan}^-^1 (0.30) =\theta[/tex] [tex]\theta = 16.69924423[/tex]The minimum angle at which the block will begin to slide is about 16.70 degrees.
A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion
[tex]{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\ [/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}[/tex]
[tex]\\[/tex]
☯ As we know that,
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }[/tex]
[tex]\\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\ [/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s = 0.04 \: m }[/tex]
[tex]\\[/tex]
☯ For left penetration (s₂)
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}[/tex]
[tex]\\[/tex]
[tex]\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\ [/tex]
An aluminum baking sheet with a mass of 225 g absorbs 2.4 x 104 J from an oven. If its temperature was initially 25 C, what will its new temperature be?
Answer:
The value is [tex]T_2 =416.9 \ K[/tex]
Explanation:
From the question we are told that
The mass of the aluminum baking sheet is [tex]m = 225 \ g = 0.225 \ kg[/tex]
The energy absorbed is [tex]E = 2.4 *10^{4} \ J[/tex]
The initial temperature is [tex]T_1 = 25 ^oC = 25 + 273 = 298 \ K[/tex]
Generally the heat absorbed is mathematically represented as
[tex]Q = m * c_a * [T_2 - T_1][/tex]
Here [tex]c_a[/tex] is the specific heat capacity of aluminum with value [tex]c_a = 897 \ J / kg \cdot K[/tex]
So
[tex]2.4 *10^{4 } =0.225 * 897 * [ T_ 2- 298][/tex]
=> [tex]T_2 - 298 = 118.915[/tex]
=> [tex]T_2 =416.9 \ K[/tex]
The speed of revolution of particle going around a circlr is doubled and its angular speed is havled. What happen to the centripetal acceleration?
a) unchanged
b) doubles
c) halves
d) becomes four times
Answer: The correct answer is C
Explanation:
g You heard the sound of a distant explosion (3.50 A/10) seconds after you saw it happen. If the temperature of the air is (15.0 B) oC, how far were you from the site of the explosion
Answer:
The answer is "1557 meters".
Explanation:
speed of sound in ([tex]\frac{m}{s}[/tex]) [tex]= 331.5 + 0.60 \ T^{\circ}\ C\\\\[/tex]
[tex]\to V = 331.5 + 0.6 \times 24 = 346 \frac{m}{s}\\\\\to t = 4.5 \ seconds \\\\\to S = vt = 346 \times 4.5 = 1557 \ meters[/tex]
which factor does not affect the strength of an electromagnet
Answer:
the placement of the ammeter in the circuit
Explanation:
A 1 200-kg automobile moving at 25 m/s has the brakes applied with a deceleration of 8.0 m/s2. How far does the car travel before it stops?
Answer:
Δx = 39.1 m
Explanation:
Assuming that deceleration keeps constant during the braking process, we can use one of the kinematics equations, as follows:[tex]v_{f} ^{2} - v_{o} ^{2} = 2* a * \Delta x (1)[/tex]
where vf is the final velocity (0 in our case), v₀ is the initial velocity
(25 m/s), a is the acceleration (-8.0 m/s²), and Δx is the distance
traveled since the brakes are applied.
Solving (1) for Δx, we have:[tex]\Delta x = \frac{-v_{o} ^{2} }{2*a} = \frac{-(25m/s)^{2}}{2*(-8.0m/s2} = 39.1 m (2)[/tex]
The car will travel a distance of 39.1 m before its stops.
To solve the problem above, use the equations of motion below.
Equation:
v² = u²+2as................... Equation 1Where:
v = final velocity of the automobileu = initial velocity of the automobilea = accelerations = distance coveredFrom the question,
Given:
v = 0 m/s (before its stops)u = 25 m/sa = -8 m/s² (decelerating)Substitute these values into equation 1
⇒ 0² = 25²+2(-8)(s)Solve for s
⇒ 0²-25² = -16s⇒ -16s = -625⇒ s = -625/16⇒ s = 39.1 mHence, The car will travel a distance of 39.1 m before its stops.
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A balloon contains 0.075 m^3 of
gas. The pressure is reduced to
100kPa and fills a box of 0.45 m^3.
What is the initial pressure inside the
balloon if the temperature remains
constant?
Answer:
600 KPa.
Explanation:
From the question given above, the following data were obtained:
Initial volume (V1) = 0.075 m³
Final volume (V2) = 0.45 m³
Final pressure (P2) = 100 KPa
Initial pressure (P1) =?
Temperature = constant
The initial pressure can be obtained by using the Boyle's law equation as shown below:
P1V1 = P2V2
P1 × 0.075 = 100 × 0.45
P1 × 0.075 = 45
Divide both side by 0.075
P1 = 45 / 0.075
P1 = 600 KPa.
Thus, the initial pressure in the balloon is 600 KPa.
A stone dropped from a bridge strikes the water 5.6 seconds later. What is the final velocity in meters/s?
A) 179.78 meters/s
B) 5.71 meters/s
C) 1.75 meters/s
D) 54.88 meters/s
Answer: 54.88 meters/s
Explanation:
The final velocity will be calculated by using the formula:
v = u + at
where,
v = final velocity
u = initial velocity = 0
a = 9.8
t = 5.6
Therefore, we slot the value back into the formula. This will be:
v = u + at
v = 0 + (9.8 × 5.6)
v = 0 + 54.88
v = 54.88 meters per second
Therefore, the final velocity is 54.88m/s
Part A:
The primary coil of a transformer contains 100 turns; the secondary has 200 turns. The primary coil is connected to a size-AA battery that supplies a constant voltage of 1.5 volts. What voltage would be measured across the secondary coil?
Part B:
A transformer is intended to decrease the value of the alternating voltage from 500 volts to 25 volts. The primary coil contains 200 turns. Find the necessary number of turns N2 in the secondary coil.
Part C:
A transformer is intended to decrease the value of the alternating current from 500 amperes to 25 amperes. The primary coil contains 200 turns. Find the necessary number of turns N2 in the secondary coil.
Part D:
In a transformer, the primary coil contains 400 turns, and the secondary coil contains 80 turns. If the primary current is 2.5 amperes, what is the secondary current I2?
Part E:
The primary coil of a transformer has 200 turns and the secondary coil has 800 turns. The power supplied to the primary coil is 400 watts. What is the power generated in the secondary coil if it is terminated by a 20-ohm resistor?
Part F:
A transformer supplies 60 watts of power to a device that is rated at 20 volts. The primary coil is connected to a 120-volt ac source. What is the current I1 in the primary coil?
Part G:
The voltage and the current in the primary coil of a nonideal transformer are 120 volts and 2.0 amperes. The voltage and the current in the secondary coil are 19.4 volts and 11.8 amperes. What is the efficiency e of the transformer? The efficiency of a transformer is defined as the ratio of the output power to the input power, expressed as a percentage: e=100Pout/Pin.
Answer:
a) 0 V
b) 10 turns
c) 4000 turns
d) 12.5 A
e) 400 W
f) 0.5 A
g) 95.4%
Explanation:
A
0
B
To solve this, we would be using the simple relationship between voltage and number of turns
V1/V2 = N1/N2
500/25 = 200/N2
20 = 200/N2
N2 = 200/20
N2 = 10 turns
C
Here also, we would be using the relationship between current and the number of turns
I1/I2 = N2/N1
500/25 = N2/20
20 = N2/20
N2 = 20 * 20
N2 = 4000 turns
D
Like in the previous question, current and the number of turn relationship is used
N1/N2 = I2/I1
400/80 = I2/2.5
5 = I2/2.5
I2 = 5 * 2.5
I2 = 12.5 A
E
The power remains unchanged at 400 W
F
Power = Voltage * Current
P = VI
I = P/V
I = 60/120
I = 0.5 A
G
95.4%
The transformer is a device used to step up or step down voltage.
Part A;
Given that;
Es/Ep = Ns/Np
Es = voltage in the secondary coil
Ep = voltage in primary coil
Ns = Number of turns in secondary coil
Np = Number of coils in primary coil
Es = Ns/Np × Ep
Es = 200/100 × 1.5 V
Es = 3 V
Part B
Ns = Es/Ep × Np
Ns = 25/500 × 200
Ns = 10 turns
Part C
Ns/Np = Ip/Is
Ns = Ip/Is × Np
Ns = 500/25 × 200
Ns = 4000 turns
Part D
Ns/Np = Ip/Is
NsIs = NpIp
Is = NpIp/Ns
Is = 400 × 2.5/80
Is =12.5 A
Part E
The power in the primary coil is the same as the power in the secondary coil. The power in the secondary coil is 400 watts.
Part F
Power supplied = 60 watts
Voltage of primary coil = 120 V
Since;
P = IV
I = P/V = 60/120 = 0.5 A
Part G
Since;
E = 100Pout/Pin
Pin = 120 V × 2 A = 240 W
Pout = 19.4 V × 11.8 A = 228.92 W
E = 100(228.92/240)
E = 95.4%
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During an experiment, Ellie records a measurement of 25,000 m. How would she write her
measurement in scientific notation?
O 2.5 x 10^2 m
O 2.5 x 10^3 m
0 2.5 x 10^5 m
O 2.5 x 10^4.m
Answer:
D. 2.5 x 10^4
Explanation:
10 to the power of 4 is 10,000.
10,000 x 2.5 = 25,000
D. 2.5 x 10⁴ is the scientific notation of Ellie records.
What is scientific notation?Scientific Notation is the expression of a number n in the form a∗10ᵇ. where a is an integer such that 1≤|a|<10. and b is an integer too. It is a way of writing very large or very small numbers. A number is written in scientific notation when a number between 1 and 10 is multiplied by a power of 10.
According to the question,
The given number is 25,000 m
i.e.
25 x 10³
Since we have to put the decimal between 2 and 5
It becomes,
2.5 x 10⁴
or
10 to the power of 4 is 10,000.
10,000 x 2.5 = 25,000
Therefore,
The answer is 2.25 x 10⁻⁶ M
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WHAT DOES DENSITY HAVE TO DO WITH PLATE TECTONICS?
Explain
Answer: The reason for the differences in density is the composition of rock in the plates. When two plates come in contact with each other through plate tectonics, scientists can use the density of the plates to predict what will happen. Whichever plate is more dense will sink, and the less dense plate will float over it.
Explanation:
Hope this helps ( not copied and pasted, this answer was done by me so I don't know if it's good or not)
A truck pushes a pile of dirt horizontally on a frictionless road with a net force of
20
N
20N20, start text, N, end text for
15.0
m
15.0m15, point, 0, start text, m, end text.
How much kinetic energy does the dirt gain?
Answer:
300 JoulesExplanation:
This is a common question on Khan Academy's "Calculating change in kinetic energy from a force" practice exercises. (AP Physics 1)
The simplest method to use is the following: [tex]W = F * d * cos(theta)[/tex], where W represents work (joules), F represents force (newtons), d represents distance (meters), and theta represents the angle of the force that's being applied.In this scenario, the force being applied is horizontal, so we can remove the [tex]cos(theta)[/tex] from our equation.So, our equation is now: [tex]W = F * d[/tex]. This would mean that [tex]W = 20 * 15[/tex], which is equal to [tex]300[/tex]. Our answer is 300 joules. (this value is positive and not negative because kinetic energy is being GAINED, not LOST)Here's the real question without all the formatting:
A truck pushes a pile of dirt horizontally on a frictionless road with a net force of 20 N for 15.0 m. How much kinetic energy does the dirt gain?A truck pushes a pile of dirt in the horizontal direction with a force of 20 N to a distance of 15 m then the kinetic energy of the dirt is 300 J.
What is work?In physics, the term "work" refers to the measurement of energy transfer that takes place when an item is moved over a distance by an externally applied, at least a portion of which is applied inside the direction of the displacement.
The duration of the path is multiplied by the component of a force acting along the path to calculate work if the forces are constant. The work W is theoretically equivalent to the pressure f times the range d, or W = fd, to represent this idea. Work is done when a force is applied at an angle of to a displacement, or W = fd cos.
The work done, [tex]W = Force * Displacement[/tex]
W = 15× 20
W = 300 J
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Question 2 of 15
The molar mass of a gas is
OA. a constant for all gases
D B. dependent on the type of gas
C. the mass of a mole of the gas
DD. dependent on the temperature
Answer:
I think it's C
Explanation:
Molar mass (M) is equal to the mass of one mole of a particular element or compound; as such, molar masses are expressed in units of grams per mole (g mol–1) and are often referred to as molecular weights. The molar mass of a particular gas is therefore equal to the mass of a single particle of that gas multiplied by Avogadro’s number (6.02 x 1023 ). To find the molar mass of a mixture of gases, you need to take into account the molar mass of each gas in the mixture, as well as their relative proportion.
which of the following is true? for this graph.
a. the object increase its velocity.
b. the object decrease its velocity.
c. the objects velocity stays unchanged.
d. the object stays at rest.
e. more information is required.
Answer:
b. The objects velocity is decreasing.
Explanation:
I know this because if you read a graph left to right it will tell u whether it is increasing or decreasing.
Can I get help on this question please
it would be the 3rd one. so C
Two clear but non-mixing liquids each of depth 15 cm are placed together in a glass container. The liquids have refractive indices of 1.75 and 1.33. What is the apparent depth of the combination when viewed from above?
Answer:
The apparent depth d = 19.8495 cm
Explanation:
The equation for apparent depth can be expressed as:
[tex]d = \dfrac{d_1} {\mu_1}+\dfrac {d_2}{\mu_2}[/tex]
here;
[tex]d_1 = d_2 = 15 \ cm[/tex]
[tex]\mu_1[/tex] = refractive index in the first liquid = 1.75
[tex]\mu_2[/tex] = refractive index in the second liqquid= 1.33
∴
[tex]d = \dfrac{15}{1.75}+\dfrac{15}{1.33}[/tex]
[tex]d = 15( \dfrac{1}{1.75}+\dfrac{1}{1.33})[/tex]
[tex]d = 15( 0.5714 +0.7519)[/tex]
d = 15(1.3233 ) cm
d = 19.8495 cm
On Venus, the atmospheric temperature is a hot 720 K due to the greenhouse effect. It consists mostly of carbon dioxide (molar mass 44 g/mol) and the pressure is 92 atm. What is the total translational kinetic energy of 3 moles of carbon dioxide molecules?
Answer:
The value is [tex]E_t = 17958.2 \ J[/tex]
Explanation:
From the question we are told that
The atmospheric temperature is [tex]T_a = 720 \ K[/tex]
The molar mass of carbon dioxide is [tex]Z = 44 \ g/mol[/tex]
The pressure is [tex]P = 92 \ atm =[/tex]
The number of moles is [tex]n = 3 \ moles[/tex]
Generally the translational kinetic energy is mathematically represented as
[tex]E_t = \frac{f}{2} * n * R T[/tex]
Here R is the gas constant with value [tex]R = 8.314 J\cdot K^{-1}\cdot mol^{-1}[/tex]
Generally the degree of freedom of carbon dioxide in terms of translational motion is f = 3
So
[tex]E_t = \frac{ 3}{2} * 2 * 8.314 * 720[/tex]
=> [tex]E_t = 17958.2 \ J[/tex]
A ball is throw at an angle of 30 degrees off the horizontal, with an initial velocity of 28 m/s. what is the maximum height the ball will reach?
[tex]{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Angle \ of \ projection = 30^{\circ} }[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ projectile = 28 \: m/s^{-1} }[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Height_{\:(max)}\: reached\: by \:the \:projectile }[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\[/tex]
☯ As we know that,
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ H = \dfrac{u^2\;sin^2\theta}{2\;g} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{H = \dfrac{(28)^2\;sin^2 30^{\circ}}{2\;(9.8)} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{H = \dfrac{784 \times \;sin^230^{\circ}}{19.6} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times sin^2 30^{\circ}}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \bigg(\dfrac{1}{2}\bigg)^2 }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \dfrac{1}{4} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: {\boxed{\sf{H=10\:m }}}[/tex]
10. A boy weighs 475 N. What is his mass? (acceleration due to gravity on Earth is 9.8m/s2 = g)
Answer: mass = 48.47 kg.
Explanation:
Formula : Weight = mg , where m = mass of body , g= acceleration due to gravity .
Given: Weight = 475 N
[tex]g= 9.8\ m/s^2[/tex]
Substitute all values in formula , we get
[tex]475= m \times9.8\\\\\Rightarrow\ m = \dfrac{475}{9.8}\\\\\Rightarrow\ m \approx 48.47\ kg[/tex]
Hence, his mass = 48.47 kg.
A speeding race car primarily contains potential energy.
:True
False
A pmdc has a stall torque of 10 and maximum mechanical power of 200. What is the maximum angular velocity?
Answer:
The maximum angular velocity is 20 rad/s
Explanation:
Given;
torque, τ = 10 N
maximum mechanical power, P = 200 J/s
The output power of the pmdc is given as;
P = τω
where;
P is the maximum mechanical power
ω is the maximum angular velocity
ω = P / τ
ω = (200) / (10)
ω = 20 rad/s
Therefore, the maximum angular velocity is 20 rad/s
20 pts
Which of the following statements is true?
1..LIghtning is a form of statlc energy.
2..Water Is a conductor of electricIty.
3Electricity must flow in a complete circuit.
4 all of the above
Answer: 2
Explanation:
:}
g A wave on a string has a speed of 13.3 m/s and a period of 0.3 s. What is the frequency of the wave
Answer:
3.33 Hz
Explanation:
The first step is to calculate the wavelength
= speed × period
= 13.3 × 0.3
= 3.99
Therefore the frequency of the wave can be calculated as follows
= speed/wavelength
= 13.3/3.99
= 3.33 Hz
why do feet smell and noses run?
Answer:
Nose has mucous glands with hairs which helps the body in trapping pollutants and infectants from entering inside the body. On the other hand,our feet is composed of millions of sweat pores when dirt and other things accumulate,it smells because of sweat mixed with the dirt and other dirty things of the ground.
Explanation:
Hope this helps
a cyclist accelerates at a rate of 7.0 m/s2. how long will it take the cyclist to go from a velocity of 4 m/s to a velocity of 18 m/s?
Answer:
2.57 seconds (rounded to 2.6 Seconds)
Step-by-step explanation:
Great question, it is always good to ask away and get rid of any doubts that you may be having.
Before we can solve this question we need to create a formula that calculates the final speed. The formula will be the following,
Where:
Vf is the final Velocity
Vi is the initial velocity
A is the acceleration
t is the time in seconds
Now that we have the formula we can plug in the values given to us in the question and solve for the amount of time (t).
Finally, we can see that it would take the cyclist 2.57 seconds (approximately) to reach a speed of 18 m/s
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
What differentiates galaxy groups from clusters?
A.
Clusters are bigger than groups.
B.
Clusters are more massive than groups.
C.
Clusters contain a hot intracluster medium, whereas groups do not.
D.
Clusters are collections of galaxy groups, whereas groups are collections of galaxies.
E.
Clusters don't gravitationally bind galaxies together, while groups bind galaxies gravitationally.
Answer:
A
Explanation:
Galaxy clusters are basically very large (>50 galaxies) groups
Answer:
The correct answer would be:
C.
Clusters contain a hot intracluster medium, whereas groups do not.
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