Answer:
297.54mExplanation:
step one:
given data
velocity v=34.2m/s
time t= 8.7s
Step two
Required is the distance the cheetah has covered on the condition
we know that speed= distance/time
make distance subject of formula we have
distance= velocity *time
distance= 34.2*8.7
distance = 297.54m
Therefore the displacement the cheetah covered at that velocity
is 297.54m
A car moves forward up a hill at 12 m/s with a uniform backward acceleration of 1.6 m/s2. What is its displacement after 6 s?
Answer:
The displacement of the car after 6s is 43.2 m
Explanation:
Given;
velocity of the car, v = 12 m/s
acceleration of the car, a = -1.6 m/s² (backward acceleration)
time of motion, t = 6 s
The displacement of the car after 6s is given by the following kinematic equation;
d = ut + ¹/₂at²
d = (12 x 6) + ¹/₂(-1.6)(6)²
d = 72 - 28.8
d = 43.2 m
Therefore, the displacement of the car after 6s is 43.2 m
A spinning ice skater will slow down if she extends her arms away from her body. Which of the following statements explain this phenomenon
A) circular motion is always uniform
B) A centripetal force always points outward
C) Angular momentum is always conserved
D) Centripetal acceleration cannot change
Marking brainliest
Answer:
B, which is why ice skaters often keep their arms close to their body when doing spins and jumps to minimize resistance.
A 5.3 kg block rests on a level surface. The coefficient of static friction is μ_s=0.67, and the coefficient of kinetic friction is μ_k= 0.48 A horizontal force, x is applied to the block. As x is increased, the block begins moving. Describe how the force of friction changes as x increases from the moment the block is at rest to when it begins moving. Show how you determined the force of friction at each of these times ― before the block starts moving, at the point it starts moving, and after it is moving. Show your work.
As the pushing force x increases, it would be opposed by the static frictional force. As x passes a certain threshold and overcomes the maximum static friction, the block will start moving and will require a smaller magnitude x to maintain opposition to the kinetic friction and keep the block moving at a constant speed. If x stays at the magnitude required to overcome static friction, the net force applied to the block will cause it to accelerate in the same direction.
Let w denote the weight of the block, n the magnitude of the normal force, x the magnitude of the pushing force, and f the magnitude of the frictional force.
The block is initially at rest, so the net force on the box in the horizontal and vertical directions is 0:
n + (-w) = 0
n = w = m g = (5.3 kg) (9.80 m/s²) = 51.94 N
The frictional force is proportional to the normal force, so that f = µ n where µ is the coefficient of static or kinetic friction. Before the block starts moving, the maximum static frictional force will be
f = 0.67 (51.94 N) ≈ 35 N
so for 0 < x < 35 N, the block remains at rest and 0 < f < 35 N as well.
The block starts moving as soon as x = 35 N, at which point f = 35 N.
At any point after the block starts moving, we have
f = 0.48 (51.94 N) ≈ 25 N
so that x = 25 N is the required force to keep the block moving at a constant speed.
As x is increasing it will be opposed by a static frictional force and for the object to start moving and maintain its acceleration, the magnitude of x must exceed the magnitude of the static frictional force and kinetic frictional force
Magnitude of normal force ( object at rest ); n = 51.94 N Required magnitude of x before the movement of object ; x = 35 NMagnitude of x after object start moving x = 25 NGiven data :
mass of block at rest ( m ) = 5.3 kg
Coefficient of static friction ( μ_s ) =0.67
Coefficient of kinetic friction is ( μ_k ) = 0.48
Horizontal force applied to block = x
First step : magnitude of normal force ( n ) when object is at rest
n = w where w = m*g
n - w = 0
n - ( 5.3 * 9.81 ) = 0 ∴ n = 51.94 N
Second step : Required magnitude of x before the movement of object
F = μ_s * n
F = 0.67 * 51.94 = 34.79 N ≈ 35 N
∴ The object will start moving once F and x = 35 N
Final step : Magnitude of x after object start moving
F = μ_k * n
= 0.48 * 51.94 = 24.93 N ≈ 25 N
∴ object will continue to accelerate at a constant speed once F and x = 25N
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A man walks south at a speed of 2.00 m/s for 60.0 minutes. He then turns around and walks north a distance 3000 m in 25.0 minutes. What is the average velocity of the man during his entire motion?
Answer:
v = 0.823 m/s
Explanation:
A man walks south at a speed of 2.00 m/s for 60.0 minutes.
The distance covered in South = 60 × 60 × 2 = 7200 m
He then turns around and walks north a distance 3000 m in 25.0 minutes.
As they moved in opposite direction, net displacement will be : 7200 - 3000 = 4200 m
Average velocity of the man = net displacement/time
[tex]v=\dfrac{4200\ m}{(60+25)\times 60}\\\\=0.823\ m/s[/tex]
So, the average velocity of the man is 0.823 m/s.
student measures the weight of a bag of bananas with a spring balance.
Describe what is inside a spring balance and explain how it works.
A spring balance measures the weight of an object by opposing the force of gravity acting with force of an extending spring. May be used to determine mass as well as weight by recalibrating the scale. Some spring balances are available in gram or kilogram markings and are used to measure the mass of an object. Spring balances consist of a cylindrical tube with a spring inside. One end (at the top) is fixed to an adjuster which can be used to calibrate the device. The other end is attached to a hook on which you can hang masses etc.
A rolling ball moves from x1 = 8.0 cm to x2 = -4.1 cm during the time from t1 = 2.9 s to t2 = 6.0 s .
Complete Question
A rolling ball moves from [tex]x_1 = 8.0 \ cm[/tex] to [tex]x_2 = - 4.1 \ cm[/tex] during the time from [tex]t_1 = 2.9 s[/tex] to [tex]t_2 = 6.0s[/tex]
What is its average velocity over this time interval?
Answer:
The velocity is [tex]v = 3.903 \ m/s[/tex]
Explanation:
From the question we are told that
The first position of the ball is [tex]x_1 = 8.0 \ cm[/tex]
The second position of the ball is [tex]x_2 = - 4.1 \ cm[/tex]
Generally the average velocity is mathematically represented as
[tex]v = \frac{ x_1 - x_2}{t_2 - t_1}[/tex]
=> [tex]v = \frac{ 8 - -4.1 }{ 6 - 2.9 }[/tex]
=> [tex]v = 3.903 \ m/s[/tex]
You serve a tennis ball of mass 60g at a speed of 50
m/s, what is the impulse exerted on the ball? ( ball starts from rest )
Answer:
[tex]J = 3~Kg.m/s[/tex]
Explanation:
Impulse and Momentum
The impulse-momentum theorem states that the change in momentum of an object equals the impulse applied to it.
The equation can be written as follows:
[tex]J =\Delta p = p_2-p_1[/tex]
Where:
J = Impulse
p2 = Final Momentum
p2 = Initial Momentum
The momentum can be calculated as:
p = m.v
Where m is the mass of the object and v is the velocity.
The tennis ball with mass m=60 g = 0.06 Kg was served from rest (v1=0) to v2=50 m/s. The change in momentum is:
[tex]\Delta p = 0.06Kg~50~m/s-0[/tex]
[tex]\Delta p = 3~Kg.m/s[/tex]
Thus the impulse is:
[tex]\marhbf{J = 3~Kg.m/s}[/tex]
Please answer my question
Answer:
Answer is (b) Mercury, venus and Mars.
Explanation:
i think b is correct!!
;-) :-) :-) :-)
In the winter sport of curling, players give a 20 kg stone a push across a sheet of ice. The Slone moves approximately 40 m before coming to rest. The final position of the stone, in principle, onlyndepends on the initial speed at which it is launched and the force of friction between the ice and the stone, but team members can use brooms to sweep the ice in front of the stone to adjust its speed and trajectory a bit; they must do this without touching the stone. Judicious sweeping can lengthen the travel of the stone by 3 m.1. A curler pushes a stone to a speed of 3.0 m/s over a time of 2.0 s. Ignoring the force of friction, how much force must the curler apply to the stone to bring it op to speed?A. 3.0 NB. 15 NC. 30 N
D. 150 N2The sweepers in a curling competition adjust the trajectory of the slope byA. Decreasing the coefficient of friction between the stone and the ice.
B. Increasing the coefficient of friction between the stone and the ice.C. Changing friction from kinetic to static.D. Changing friction from static to kinetic.3. Suppose the stone is launched with a speed of 3 m/s and travel s 40 m before coming to rest. What is the approximate magnitude of the friction force on the stone?A. 0 NB. 2 NC. 20 ND. 200 N4. Suppose the stone's mass is increased to 40 kg, but it is launched at the same 3 m/s. Which one of the following is true?A. The stone would now travel a longer distance before coming to rest.B. The stone would now travel a shorter distance before coming to rest.C. The coefficient of friction would now be greater.D. The force of friction would now be greater.
Answer:82. Since you have a distance and a force, then the easiest principle to use is energy, i.e. work.
The work done by friction is F * d. This work cancels out the kinetic energy of the stone (1/2)mv^2
Fd = (1/2)mv^2
F = (1/2)mv^2/d.
Plug in m = 20 kg, v = 3 m/sec, d = 40 m.
83. With more mass, the kinetic energy is higher now. The work needed is higher. W = F * d and F is the same.
Explanation:Hope I helped :)
A projector lens projects an image from a 6.35 cm wide LCD screen onto a
screen 3.25 m wide. If the focal length of the projector lens is 13.8 cm, the screen
must be how far from the projector
Answer:
For any given projector, the width of the image (W) relative to the throw distance (D) is know as the throw ratio D/W or distance over width. So for example, the most common projector throw ratio is 2.0. This means that for each foot of image width, the projector needs to be 2 feet away or D/W = 2/1 = 2.0.
When particles get close to the surface, they interact with atoms in
the
(Finish the sentence)
A projectile is shot straight up from the earth's surface at a speed of 11,000 km/hr. How high does it go? ________km?
Taken from "Physics for Scientists and Engineers by Randall D. Knight 2nd Edition. Chapter 13 #34. There is an answer in the database already, but I do not understand it.
Answer:
476.35 km
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 11000 km/hr
Final velocity (v) = 0 km/hr (at maximum height)
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h) = ?
Next, we shall convert 9.8 m/s² to km/hr². This is illustrated below:
1 m/s² = 12960 km/hr²
Therefore,
9.8 m/s² = 9.8 m/s² × 12960 km/hr² / 1 m/s²
9.8 m/s² = 127008 km/hr²
Thus, 9.8 m/s² is equivalent to 127008 km/h²
Finally, we shall determine the maximum height reached by the projectile.
This is illustrated below:
Initial velocity (u) = 11000 km/hr
Final velocity (v) = 0 km/hr (at maximum height)
Acceleration due to gravity (g) = 127008 km/hr²
Maximum height (h) = ?
v² = u² – 2gh (since the projectile is going against gravity)
0² = 11000² – (2 × 127008 × h)
0 = 121×10⁶ – 254016h
Collect like terms
0 – 121×10⁶ = – 254016h
– 121×10⁶ = – 254016h
Divide both side by – 254016
h = – 121×10⁶ / – 254016
h = 476.35 km
Thus, the maximum height reached by the projectile is 476.35 km
What would happen if there is more male hyenas than female hyenas in a population?
Choices:
Male hyenas will compete to mate with the females.
Some male hyenas will die.
Male hyenas for wait for more females to join the population.
Answer:
Option 1
Explanation:
I always see animals do that
A baseball is thrown across the field. The ____________is measured from where the ball is thrown to where landed was 75 feet.
motion
direction
distance
reference point
Answer:
distance i think
Explanation:
The earliest mineral observed to showmagnetic properties is called
A leadstone
Blodestone
Cloadstone
Dnone of the above
E all of the above
Answer:
B: lodestone
Explanation:
Each magnet has its magnetic poles, north (N) and south (S). Diversified ones are attracted and reptiles of the same name are repelled, similarly to charges, so it was considered possible to separate the magnet at the north and south poles.
Magnetic properties can be lost if the magnet is exposed to high temperatures if it falls or due to some mechanical shocks.
While riding a multispeed bicycle, the rider can select the radius of the rear sprocket that is fixed to the rear axle. The front sprocket of a bicycle has radius 12.0 cm. If the angular speed of the front sprocket is 0.600 rev/s, what is the radius of the rear sprocket for which the tangential speed of a point on the rim of the rear wheel will be 5.00 m/s?
Answer:
2.9 cm
Explanation:
Assuming that the rear wheel has a radius of 0.330 m
Given that
r(a) = 12 cm -> 0.12 m
w(a) = 0.6 rev/s -> 3.77 rad/s
v = 5 m/s
r(w) = 0.330 m
The speed on any point on the rim at the sprocket in the front is
v(a) = w(a).r(a) = 3.77 * 0.12 = 0.4524 m/s
Also,
v(a) = speed at any point on the chain
v(b) = speed at any point on the rim of the rear sprocket
v(a) = v(b)
where v(b) = w(b).r(b)
Recall that the speed at any point on the rear wheel is v, where
v = w(b).r(w)
5 = w(b) * 0.330
w(b) = 5/0.330
w(b) = 15.15 rad/s
On substituting this in the equation, we have
v(b) = w(b).r(b).
Remember also, that v(a) = v(b), so
0.4524 = 15.15 * r(b)
r(b) = 0.4524 / 15.15
r(b) = 0.029 m -> 2.9 cm
Therefore, the radius of the rear sprocket needed is 2.9 cm
A large pizza is cut into 8 even slices. A person orders 4 large pizzas from a restaurant. How many total slices of pizza did the person order?
Answer:
32 slicesExplanation:
Step one:
given data
we are told that 1 large pizza can be cut into 8 even slices
Required
we want to find how many slices are there in 4 large pizzas
Step two:
so if 1 pizza has 8 slices
4 pizza will have x
cross multiply we have
x= 8*4
x=32 slices
If the particles were moving with a speed much less than c, the magnitude of the momentum of the second particle would be twice that of the first. However, what is the ratio of the magnitudes of momentum for these relativistic particles?
Answer:
p₂ / p₁ = 2 (v₁ / v₂)
Explanation:
The moment is a very useful concept, since it is one of the quantities that is conserved during shocks and explosions, for which it had to be redefined to be consistent with special relativity,
p = m v / √[1+ (v/c)² ]
for the case of speeds much lower than the speed of light this expression is close to
p = m v
In this exercise they indicate that the moment of the second particle is twice the moment of the first, when their velocities are small
p₂ = 2 p₁
p₂/p₁ = 2
in consecuense
m v₂ = 2 m v₁
v₂ = 2 v₁
consider particles of equal mass.
By the time their speeds increase they enter the relativistic regime
p₂ = mv₂ /√(1 + v₂² /c²)
p₁ = m v₁ /√(1 + v₁² / c²)
let's look for the relationship between these two moments
p₂ / p₁ = mv₂ / mv₁ [√ (1+ v₁² / c²) /√ (1 + v₂² / c²)
from the initial statement
p₂ / p₁ = 2 √(c² + v₁²) / (c² + v₂²)
we take c from the root
p₂ / p₁ = 2 √ [(1+ v₁²) / (1 + v₂²)]
this is the exact result, to have an approximate shape suppose that the velocities are much greater than 1
p₂ / p₁ = 2 √ [v₁² / v₂²] = 2 √ [(v₁ / v₂)²]
p₂ / p₁ = 2 (v₁ / v₂)
we see the value of the moment depends on the speed of the particles
7. Answer each of the following questions if the student applied a 550N force to a 100kg box on the same surface. What would be the magnitude of the box's acceleration?
Acceleration of an object is the force divided by its mass. Hence, acceleration of the 100 Kg box with an applied force of 550 N is 5.5 m/s².
What is acceleration?Acceleration is a physical quantity having both magnitude and direction. This vector measures the rate of change of velocity of a moving body. Thus, acceleration has the unit of m/s².
According to Newton's second law of motion, force acting on a body is the product of its mass and acceleration. Thus, force is directly proportional to the mass and acceleration of the body.
Given the mass of the box = 100 Kg
Force applied on it = 550 N.
Acceleration = 550 N / 100 Kg
= 5.5 m/s² ( 1 N = 1 Kg m/s²)
Therefore, the acceleration of the box is 5.5 m/s².
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during a baseball game you are running home and slide into home plate. However you come up short and you are tagged out. Which force stops you from sliding all the way home? a friction b gravity c pull d push
Answer:1 because
Explanation: it’s pointing to the earth and gravity
Pulls things down to earth
2. Which bicyclist was traveling the fastest at the end of the race?
Answer:
This question is incomplete
Explanation:
This question is incomplete. However, to determine the bicyclist that traveled the fastest at the end of the race, the speed of the bicyclists at the end of the race will determine this (not the bicyclist that came first nor there overall speed). The speed of the bicyclist at the end of the race can be determined by using the formula below
s = d ÷ t
Where s is the speed of each bicyclist at the end of the race
d is the specific distance covered by the bicyclist at the end of the race
t is the time taken for the bicyclist to complete that distance
It should be noted that to get an accurate result, the distance covered at the end of the race must be the same for all the bicyclists.
A ball is thrown vertically upward with an initial velocity of 23 m/s. What are its position and velocity after 2 s?
Answer:
The position of the ball after 2 s is 26.4 mThe velocity of the ball after 2 s is 3.4 m/sExplanation:
Given;
initial velocity of the ball, u = 23 m/s
time of motion, t = 2 s
The position of the ball after 2 s is given by;
h = ut - ¹/₂gt²
h = (23 x 2) - ¹/₂ x 9.8 x 2²
h = 46 - 19.6
h = 26.4 m
The velocity of the ball after 2 s is given by;
v² = u² + 2(-g)h
v² = u² - 2gh
v² = 23² - (2 x 9.8 x 26.4)
v² = 529 - 517.44
v² = 11.56
v = √11.56
v = 3.4 m/s
Acceleration is sometimes expressed in multiples of g, where g = 9.8 m/s^2 is the magnitude of the acceleration due to the earth's gravity. In a test crash, a car's velocity goes from 26 m/s to 0 m/s in 0.15 s. How many g's would be experienced by a driver under the same conditions?
Answer:
Acceleration = 18g
Explanation:
Given the following data;
Initial velocity, u = 26m/s
Final velocity, v = 0
Time = 0.15 secs
To find the acceleration;
In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.
This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.
Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.
Mathematically, acceleration is given by the equation;
[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]
Substituting into the equation, we have;
[tex]a = \frac{0 - 26}{0.15}[/tex]
[tex]a = \frac{26}{0.15}[/tex]
Acceleration = 173.33m/s2
To express it in magnitude of g;
Acceleration = 173.33/9.8
Acceleration = 17.7 ≈ 18g
Acceleration = 18g
How much work would be done on a particle with 5.0 C of charge on it if it moved from an equipotential line at 5.5 volts to another equipotential line at 3.5 volts?
Answer:
10J
Explanation:
In this question we have the following information
The charge of the particle is q = 5 C
The equipotenetial level is V1 = 5.5 v
and also the
equipotenetial level is V2 = 3.5 v
So we calculate the
work done W=q x (v1-v2)
workdone = 5 x (5.5-3.5)
= 5x2
=10 J
Workdone = 10 J
So we conclude that the workdone on a particle with these information is 10j
A racecar accelerates from rest at 6.5 m/s2 for 4.1 s. How fast will it be going at the end of that time?
Answer:
The final velocity of the car is 26.65 m/s.
Explanation:
Given;
acceleration of the racecar, a = 6.5 m/s²
initial velocity of the car, u = 0
time of motion, t = 4.1 s
The final velocity of the car is given by;
v = u + at
where;
v is the final velocity of the car
suvstitute the givens
v = 0 + (6.5)(4.1)
v = 26.65 m/s.
Therefore, the final velocity of the car is 26.65 m/s.
"2.40 A pressure of 4 × 106N/m2 is applied to a body of water that initially filled a 4300 cm3 volume. Estimate its volume after the pressure is applied."
Answer:Final volume after pressure is applied=4,292cm3
Explanation:
Using the bulk modulus formulae
We have that The bulk modulus of waTer is given as
K =-V dP/dV
Where K, the bulk modulus of water = 2.15 x 10^9N/m^2
2.15 x 10^9N/m^2= - 4,300 x 4 × 106N/m2 / dV
dV = - 4,300 x 4 × 10^6N/m^2/ 2.15 x 10^9N/m^2
dV (change in volume)= -8.000cm^3
Final volume after pressure is applied,
V= V+ dV
V= 4300cm3 + (-8.000cm3)
=4300cm3 - 8.000cm3
Final Volume, V =4,292cm3
Bob rides his bike with a constant speed of 10 miles per hour. How long will he take to travel a distance of 15 miles?
[tex]{\underline{\pink{\textsf{\textbf{ Answer : }}}}}[/tex]
➡ 150hrs.
[tex]{\underline{\pink{\textsf{\textbf{Explanation : }}}}}[/tex]
➡ Time = distance × speed
➡ Time = 15*10
➡ Time = 150hrs ans.
What is the current in the wire now?
Answer:
220v
Explanation:
Sorry, the question is incomplete
Answer:
on the potential difference applied and on the resistance of the wire.
Explanation:
Ohms law state that the current through a conductor between two points is directly proportional to the potential difference across the two points. Imtroducing the comstant of proportionality, the resistance, one arrives at the usual athematical equation that describes this relationship: I = V/R.
A circular conducting loop with a radius of 1.00 m and a small gap filled with a 10.0 Ω resistor is oriented in the xy-plane. If a magnetic field of 2.0 T, making an angle of 30º with the z-axis, increases to 11.0 T, in 2.5 s, what is the magnitude of the current that will be caused to flow in the conductor?
Answer:
ill get back to this question once i find the answer to it
8. A rectangle is measured to be 6.4 +0.2 cm by 8.3 $0.2 cm.
a) Calculate its perimeter in cm
b) Calculate the uncertainty in its perimeter.
Answer:
a) The perimeter of the rectangle is 29.4 centimeters.
b) The uncertainty in its perimeter is 0.8 centimeters.
Explanation:
a) From Geometry we remember that the perimeter of the rectangle ([tex]p[/tex]), measured in centimeters, is represented by the following formula:
[tex]p = 2\cdot (w+l)[/tex] (1)
Where:
[tex]w[/tex] - Width, measured in centimeters.
[tex]l[/tex] - Length, measured in centimeters.
If we know that [tex]w = 6.4\,cm[/tex] and [tex]l = 8.3\,cm[/tex], then the perimeter of the rectangle is:
[tex]p = 2\cdot (6.4\,cm+8.3\,cm)[/tex]
[tex]p = 29.4\,cm[/tex]
The perimeter of the rectangle is 29.4 centimeters.
b) The uncertainty of the perimeter ([tex]\Delta p[/tex]), measured in centimeters, is estimated by differences. That is:
[tex]\Delta p = 2\cdot (\Delta w + \Delta l)[/tex] (2)
Where:
[tex]\Delta w[/tex] - Uncertainty in width, measured in centimeters.
[tex]\Delta l[/tex] - Uncertainty in length, measured in centimeters.
If we know that [tex]\Delta w = 0.2\,cm[/tex] and [tex]\Delta l = 0.2\,cm[/tex], then the uncertainty in perimeter is:
[tex]\Delta p = 2\cdot (0.2\,cm+0.2\,cm)[/tex]
[tex]\Delta p = 0.8\,cm[/tex]
The uncertainty in its perimeter is 0.8 centimeters.