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Ivanshal [37]
4 years ago
8

Whats 6 3/7 ×1 5/9.

Physics
2 answers:
aniked [119]4 years ago
7 0
6 3/7 * 1 5/9
45/7 * 14/9
630/63
10
tatyana61 [14]4 years ago
6 0
That would most likely be 10
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The average force applied to move an initially stationary golf cart over a period of 25 s is 500 N. What is the change in moment
Yuki888 [10]

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12500(kg*m/s)

Explanation:

F=ma=mv/t=p/t

p=F*t=500N*25 s=12500(kg*m/s)

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3 years ago
How the energy is transformed
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motion energy to sound energy

Explanation:

you move your hands together, then clap then, which makes a sound. Hope that this helps you and have a great day :)

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Cells are considered the smallest part of an organsim that can?​
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That can be considered alive.
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3 years ago
Check all of the following statements that describe velocity.
Nata [24]

Answer:

Choices A, B, and C are correct.

Explanation:

Let us look at each of the choices one by one:

A. It is a vector

Yes. Velocity is a vector, or it's a speed with direction.

B. It is the change in displacement divided by the change in time.

Yes. The velocity can be written as

v = \dfrac{\Delta\bold{x}}{\Delta t}

where \bold{x} is the displacement—a vector quantity.

C. It can be measured in meters per second.

Yes. The units of velocity are m/s, but also with a unit vector indicating the direction.

D. It is the slope of the acceleration vs. time graph.

Nope. The velocity is the slope of displacement vs. time graph.

Hence, only choices A, B, and C are correct.

4 0
3 years ago
A 3.9 g dart is fired into a block of wood with a mass of 24.6 g. The wood block is initially at rest on a 1.5 m tall post. Afte
Galina-37 [17]

Answer:

46.48m/s

Explanation:

The problem is a combination of the principle of conservation of linear momentum and projectile motion.

The principle of conservation of linear momentum states that in a closed system, the total momentum of colliding bodies before impact is equal to the total momentum after impact. The masses stated in the problem experienced an inelastic collision. In an inelastic collision, the bodies involved stick together after the collision and move with a common velocity.

For two bodies of masses m_1 and m_2 moving with velocities u_1 and u_2 before impact, if they experience inelastic collision, the conservation of their momenta is as stated in equation (1);

m_1u_1+m_2u_2=(m_1+m_2)v..................(1)

were v is their common velocity after impact. If the second mass m_2 was at rest before the impact, then its initial velocity u_2=0m/s. therefore m_2u_2=0. Equation (1) then becomes;

m_1u_1=(m_1+m_2)v..............(2)

In the problem stated, the second mass taken as the mass of the wooden block was at rest before the impact and the collision was inelastic since both the wood and the dart stuck together and moved with a common velocity after the impact. Therefore we can use equation (2) for the problem.

Given;

m_1=3.9g=0.0039kg\\u_1=?\\m_2=24.6g=0.0246kg\\v=?

Substituting these values into (2), we get the following;

0.0039*u_1=(0.0039+0.024)v\\0.0039u_1=0.0285v.........(3)

Their common v velocity after impact now makes both the wooden block and the dart (as a single body) to fall vertically through a height h of 1.5m over a range R of 3.5m as stated by the problem; hence by the principle of projectile motion for a body projected horizontally, the following relationship holds;

R= vt............(4)

were t is the time taken to fall through the height h. To obtain t we use the second equation of free fall under gravity;

h=\frac{1}{2}gt^2...........(5)

were g is acceleration due to gravity taken as 9.8m/s^2. Therefore;

1.5=\frac{1}{2}*9.8*t^2\\1.5=4.9t^2\\t^2=\frac{1.5}{4.9}=0.306\\t=\sqrt{0.306} =0.55s

We then substitute R and t into equation (4) to obtain v.

3.5=v*0.55\\v=\frac{3.5}{0.55}\\v=6.36m/s

We now further substitute this value of v into (3) to obtain u_1;

u_1=\frac{0.0285v}{0.0039}\\\\u_1=\frac{0.0285*6.36}{0.0039}\\\\u_1=\frac{0.18126}{0.0039}\\\\u_1=46.48m/s

4 0
3 years ago
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