Terry is walking down the street at 3 m/s. If he
has a mass of 70 kg, what is his momentum?

The magnitude of the average force exerted on the ball by the club is 429.27 N.

Force can be defined as the product of mass and acceleration

To calculate the force exerted on the ball by the club, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, The magnitude of the average force exerted on the ball by the club is 429.27 N.

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Answer:

bye bye

Explanation:

Answer: Or please call again to see if they reached the phone yet.

Explanation: It always stops ringing right before someone gets to there phone so call again in 2 minutes. :3

The two components of projectile motion include the following:

This is defined as a missile propelled by the application of an external force and allowed to move freely under the influence of gravity and air resistance.

The equation for Horizontal motion Vx = V * cos(α)

Vertical velocity component: Vy = V * sin(α)

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Answer:

0.98 [sec].

Explanation:

1) the time when the player is moving up equals the time the player falls, it can be 't'. Then

2) the equation of moving down can be written as

[tex]\frac{g*t^2}{2} =h, \ where[/tex]

g=10 [m/s²], t - time, h - the given height.

3) according to the formula above the time spent to moving down is:

[tex]t=\sqrt{\frac{2h}{g}};[/tex]

[tex]t=\sqrt{\frac{2.4}{10}}=\sqrt{0.24} =0.49[s].[/tex]

4) finally, the total time is t*2=0.98[sec].

Answer:

The answer is B. Friction on the roller blades is greater on asphalt than on concrete.

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Answer:

See below.

Explanation:

According to the question, we know that,

work done is given by,  [tex]W=qV[/tex]

and change in kinetic energy is, Δ [tex]KE=W=1=1/2[mv^{2} ][/tex]

therefore equating both the equations we get,

[tex]qV=1/2[mv^{2} ][/tex] ⇒ [tex]V=\frac{mv^{2} }{2q}[/tex]

m= mass of electron =  [tex]9.1*10^{-31} kg[/tex]

q= charge on an electron = [tex]1.6*10^{-19} C[/tex]

v= speed of electron= 700000m/s

substituting the values in the above equation, we get

[tex]V=\frac{9.1*10^{-31} *(700000)^{2} }{2*1.6*10^{-19} } =1.39V[/tex]

(1).  the potential difference that stopped the electron is 1.39 volts.

now the kinetic energy equation is :  2 ways[tex]KE=1/2[mv^{2} ]=\frac{9.1*10^{-31} *700000^{2} }{2} =2.22*10^{-19} J\\[/tex]

or [tex]KE=\frac{2.22*10^{-19} }{1.6*10^{-19} } =1.39eV[/tex]

(2).  the initial kinetic energy of the electron is 1.39eV.

Answer:

M1 would seem to be slower because of a larger mass

x1 = A1 sin ω1 t1        describes the displacement

ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2  since k's are equal

ω1 / ω2 = 1/2 from graph    (frequency of 2 is greater)

(m1 / m2)^1/2 = ω2 / ω1    from above

m1 / m2 = 2^2 = 4    so m1 would have 4 times the mass of m2

M1 would seem to be slower because of a larger mass

x1 = A1 sin ω1 t1      

ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2  since k's are equal

ω1 / ω2 = 1/2 from graph   (frequency of 2 is greater)

(m1 / m2)^1/2 = ω2 / ω1   from above

m1 / m2 = 2^2 = 4    so m1 would have 4 times the mass of m2.

The two graphs shown represent the motion of two blocks with different masses, m1 and m2. The blocks are oscillating on identical springs. For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system. For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.

The two graphs shown represent the motion of two blocks with different masses, m1 and m2. The blocks are oscillating on identical springs. For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

Therefore, M1 would seem to be slower because of a larger mass

x1 = A1 sin ω1 t1      

ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2  since k's are equal

ω1 / ω2 = 1/2 from graph   (frequency of 2 is greater)

(m1 / m2)^1/2 = ω2 / ω1   from above

m1 / m2 = 2^2 = 4    so m1 would have 4 times the mass of m2.

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Answer:

A. Gravity pulls on the book :)

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Answer:

Lithium iodide

Explanation:

it's formula is LiI

Lithium iodide is a chemical compound composed of lithium and iodine, which when exposed to air turns yellow due to the oxidation of iodide to iodine. It has a chemical formula LiI.

The products, lithium iodide (LiI), sodium iodide (NaI), and cesium iodide (CsI) look like typical ionic compounds; they are all white crystalline solids

Answer:

Power, technique, and timing

Explanation:

I took the test, hope it helps, dont forget to rate ⭐⭐⭐⭐

Answer:

A

Explanation:

I took the quiz

It is amount of force or motion that cause the movement or moving of body.

Thus it is well explained.

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1.No they do not violate the conservation of law. Because the initial momentum can be zero.

2. The momentum is conserved in both elastic and inelastic collision.

The momentum of the body before the contact is always equal to the momentum of the body after the impact, according to the rule of conservation of momentum.

According to the law of conservation of momentum;

Momentum before collision =Momentum after collision

[tex]\rm m_1u_1 + m_2u_2 = v(m_1 + m_2)[/tex]

1.No they do not violate the conservation of law. Because the initial momentum can be zero.

2. The momentum is conserved in both elastic and inelastic collision. Hence the conservation law works with elastic and inelastic collisions.

3. An elastic collision is one in which kinetic energy is conserved, meaning that the total kinetic energy before and after the impact is the same.

The object bounces once after the body gets struck. When the body's final velocity is zero it follows the stationary object after the collision.

4.. Airbags are meant to inflate automatically in the case of an abrupt deceleration or impact force indicative of a collision.

When the bag is fully inflated, it protects the driver and/or passenger by Increasing cushioning around the driver's neck, head, and spine.

5. Stretchable seatbelts are meant to allow a passenger's considerable velocity to be slowly reduced, preventing the passenger from being flung forward abruptly.

As a result, if the automobile collides, serious casualties can be avoided.

Hence no they do not violate the conservation of law. Because the initial momentum can be zero.

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The amplitude of the wave on the given sinusoidal wave graph is 10 cm.

The amplitude of a wave is the maximum displacement of a wave. This is the highest vertical position of the wave from the origin.

Amplitude of the wave is calculated as follows;

From the graph, the amplitude of the wave or maximum displacement of the wave is 10 cm.

Thus, the amplitude of the wave on the given sinusoidal wave graph is 10 cm.

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This question involves the concept of  Newton's law of gravitation.

The force of gravity between the two spaceships is "1355.78 N".

According to Newton's Law of Gravitation:

[tex]F=\frac{Gm_1m_2}{r^2}[/tex]

where,

Therefore,

[tex]F=\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(500\ kg)(498\ kg)}{(35\ m)^2}\\\\[/tex]

F = 1355.78 N = 1.356 KN

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Two space ships traveling next to each other. The first one is 500 kg and the second one is 498 kg. They are 35 meters apart, the Force of gravity between the two spaceships is 1355.78 N.

It is given that the First spaceship's weight ([tex]m_{1}[/tex]) is 500 kg,

The second spaceship's weight ([tex]\rm m_{2}[/tex]) is 498 kg.

The distance between spaceships (r) is 35 meters.

It is required to find the Force of gravity between these spaceships.

It is defined as the force which attracts any two masses in the universe.

By Newton's law of Gravitation:

[tex]\rm F= \frac{Gm_1m_2}{r^2}[/tex]  , Where

[tex]\rm F = The\ force \ of \ gravity \ between \ the \ spaceships\\\rm G= Universal\ Gravitational \ Constant = 6.67 \times 10^{-11} N.m^2/kg^2[/tex]

Putting values in the above formula:

[tex]\rm F = \frac{(6.67\times 10^{-11} N.m^2/kg^2)(500kg)(498kg)}{(35m)^2}[/tex]

F = 1355.78 N = 1.356  KN

Thus, Two spaceships travel next to each other. The first one is 500 kg and the second one is 498 kg. They are 35 meters apart, the Force of gravity between the two spaceships is 1355.78 N.

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Answer: b or d

Explanation: b or d

Answer:

a

Explanation:

F= ma

interestingly

when you increase the mass the acceleration decreases while when the mass decreases the acceleration increases

(man, PHYSICS IS JUST THE BESY)

A: a

object with the smallest mass has largest acceleration

The statements that are true are;

Work is said to be done when te force applied travels a distance in the direction of the force. Now we can see that when the force is applied by shoving the crate, work is done, an additional work is done by friction to bring the crate to a stop.

Hence, the following are true;

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i want the answer of this exercise

Answer:

50 N

Explanation:

Let the natural length of the spring = L

so

100 = k(40 - L)       (1)

200 = k(60 - L)       (2)

(2)/(1):   2 = (60 - L)/(40 - L)

60 - L = 2(40 - L)

60 - L = 80 - 2L

2L - L = 80 - 60

L = 20

Sub it into (1):

100 = k(40 - 20) = 20k

k = 100/20 = 5 N/in

Now

X = k(30 - L) = 5(30 - 20) = 50 N

The electric field is all around the balloon and paper including between them, where, depending on their charge, there would be a force of either attraction or repulsion.

An electric field is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them.

Thus, for the charged balloon and a charged piece of paper 0.04 m away, we can conclude that, the electric field is all around the balloon and paper including between them, where, depending on their charge, there would be a force of either attraction or repulsion.

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Answer:

Approximately [tex]1.1 \times 10^{4}\; {\rm m\cdot s^{-1}}[/tex] if air friction is negligible.

Explanation:

Let [tex]G[/tex] denote the gravitational cosntant. Let [tex]M[/tex] denote the mass of the earth. Lookup the value of both values: [tex]G \approx 6.67 \times 10^{-11}\; {\rm N\cdot m^{2}\cdot kg^{-2}}[/tex] while [tex]M \approx 5.697 \times 10^{24}\; {\rm kg}[/tex].

Let [tex]m[/tex] denote the mass of the meteor.

Let [tex]v_{0}[/tex] denote the initial velocity of the meteor. Let [tex]r_{0}[/tex] denote the initial distance between the meteor and the center of the earth.

Let [tex]r_{1}[/tex] denote the distance between the meteor and the center of the earth just before the meteor lands.

Let [tex]v_{1}[/tex] denote the velocity of the meteor just before landing.

The radius of planet earth is approximately [tex]6.371 \times 10^{6}\; {\rm m}[/tex]. Therefore:

Note the significant difference between the two distances. Thus, the gravitational field strength (and hence acceleration of the meteor) would likely have changed significant during the descent. Thus, SUVAT equations would not be appropriate.

During the descent, gravitational potential energy ([tex]\text{GPE}[/tex]) of the meteor was turned into the kinetic energy ([tex]\text{KE}[/tex]) of the meteor. Make use of conservation of energy to find the velocity of the meteor just before landing.

Initial [tex]\text{KE}[/tex] of the meteor:

[tex]\displaystyle (\text{Initial KE}) = \frac{1}{2}\, m\, {v_{0}}^{2}[/tex].

Initial [tex]\text{GPE}[/tex] of the meteor:

[tex]\displaystyle (\text{Initial GPE}) &= -\frac{G\, M\, m}{r_{0}}[/tex].

(Note the negative sign in front of the fraction.)

Just before landing, the [tex]\text{KE}[/tex] and the [tex]\text{GPE}[/tex] of this meteor would be:

[tex]\displaystyle (\text{Final KE}) = \frac{1}{2}\, m\, {v_{1}}^{2}[/tex].

[tex]\displaystyle (\text{Final GPE}) &= -\frac{G\, M\, m}{r_{1}}[/tex].
If the air friction on this meteor is negligible, then by the conservation of mechanical energy:

[tex]\begin{aligned}& (\text{Initial KE}) + (\text{Initial GPE}) \\ =\; & (\text{Final KE}) + (\text{Final GPE})\end{aligned}[/tex].

[tex]\begin{aligned}& \frac{1}{2}\, m\, {v_{0}}^{2} - \frac{G\, M\, m}{r_{0}} \\ =\; & \frac{1}{2}\, m\, {v_{1}}^{2} - \frac{G\, M\, m}{r_{1}}\end{aligned}[/tex].

Rearrange and solve for [tex]v_{1}[/tex], the velocity of the meteor just before landing:

[tex]\begin{aligned}{v_{1}} &= \sqrt{\frac{\displaystyle \frac{1}{2}\, m\, {v_{0}}^{2} - \frac{G\, M\, m}{r_{0}} + \frac{G\, M\, m}{r_{1}}}{(1/2)\, m}} \\ &= \sqrt{{v_{0}}^{2} - \frac{G\, M}{r_{0}} + \frac{G\, M}{r_{1}}} \\ &= \sqrt{{v_{0}}^{2} - G\, M\, \left(\frac{1}{r_{1}} - \frac{1}{r_{0}}\right)}\end{aligned}[/tex].

Substitute in the values and evaluate:

[tex]\begin{aligned}v_{1} &= \sqrt{{v_{0}}^{2} - G\, M\, \left(\frac{1}{r_{1}} - \frac{1}{r_{0}}\right)} \\ &\approx \sqrt{\begin{aligned}(& 6.5 \times 10^{3}\; {\rm m \cdot s^{-1}}) \\ & - [6.67 \times 10^{-11}\; {\rm N \cdot {m}^{2}\cdot {kg}^{2} \times 5.697\; {\rm kg}}\\ &\quad\quad \times (1 / (6.371 \times 10^{6}\; {\rm m}) - 1 / (1.9371 \times 10^{7}\; {\rm m}))]\end{aligned}} \\ &\approx 1.1 \times 10^{4}\; {\rm m\cdot {s}^{-1}}\end{aligned}[/tex].

(Note that assuming a constant acceleration of [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] would give [tex]v_{1} \approx 1.7\times 10^{4}\; {\rm m\cdot s^{-1}}[/tex], an inaccurate approximation.

(a) The velocity (in m/s) of the rock after 1 second is 11.28 m/s.

(b) The velocity of the rock after 2 seconds is 7.56 m/s.

(c) The time for the block to hit the surface is 4.03.

(d) The velocity of the block at the maximum height is 0.

The velocity of the rock is determined as shown below;

Height of the rock after 1 second; H(t) = 15(1) - 1.86(1)² = 13.14 m

v² = u² - 2gh

where;

v² = (15)² - 2(3.72)(13.14)

v² = 127.23

v = √127.23

v = 11.28 m/s

v = dh/dt

v = 15 - 3.72t

v(2) = 15 - 3.72(2)

v(2) = 7.56 m/s

dh/dt = 0

15 - 3.72t = 0

t = 4.03 s

v = u - gt

v = 15 - 3.72(4.03)

v = 0

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Answer:

0.175 second

Explanation:

i hope it helps

Answer:

I don't understand your question ❓,the object.....of what experiment

Answer:

let alpha be y,

1/2=y(x-6)^(-b)

1/2=y(1/x-6)

1/2=y/x-6

x-6=2y

x-6/2=y

(1/2)x-(6/2)=y

y=(1/2)x-3

Answer:

2.3×10^-16

Explanation:

Charge of proton = 1.6×10^-19

let q(a) and q(b) be the two protons.

According to the Coulumbs Laws ;

F = kq(a)×q(b) / r^2

where k = 9.0×10^9 and r is the distance of seperation.

So having a distance of 1.0×10^-6 in our case ;

F = (9.0×10^9×1.6×10^-19×1.6×10^-19)/(1.0××10^-6)^2

F = (2.304×10^-28)/(1.0×10^-12)

F = 2.304×10^-16N

Since the both charges are positively charged they will repel .

Answer:

approximately 7 seconds

Explanation:

The given scenario is just an example of Projectile Motion.

In this question, we are given:

- Initial Velocity (u) = 40 m/s

- Angle of Elevation (θ) = 60 degrees

- Distance of Bird (s) = 145 meters        [Not used for the time of flight]

Time of Flight:

[tex]T = \frac{2uSin(\theta)}{g}[/tex]

we will use g = 9.8 m/s/s since we are not told otherwise.
[tex]T = \frac{2(40)Sin(60 degrees)}{9.8}[/tex]

[tex]T = \frac{40\sqrt{3}}{9.8}[/tex]

T = 4.081(√3) seconds
rounded to 7 seconds

Answer:

Explanation:

Because if you don't have strong bones, you will have problems with your health. So, a strong skeletal system is important for leading a healthy life.

Answer:

Adam (Arabic:, romanized:dam) is said to be the first human person on Earth, as well as the first prophet (Arabic:, nab) of Islam. Muslims see Adam's role as the father of the human race with veneration.

Explanation:

Answer:

Adam is the name given to the first person in Genesis 1-5. Aside from being the name of the first man, Adam is also used as a pronoun in the Bible, both individually as "a human" and collectively as "mankind."

Explanation:

Answer:The Answer is economic gain

Explanation:

Answer: 6067.5 N

Explanation:

Work = Change in Energy. To start, all of the energy is kinetic energy, so find the total KE using: KE = 1/2(m)(v^2). Plug in 1980 kg for m and 15.5 m/s for v and get KE = 237847.5 J.

Now, plug this in for work: Work = Force * Distance; so, divide work by distance to get 6067.5 N.