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stira [4]
2 years ago
8

1) Write Newton's three laws of motion (20 points). 2) Define acceleration (20 points). 3) Write the Month and Day of the Assign

ment Due Date. For example, if the assignment is due on July 15th, then Month
Physics
1 answer:
vaieri [72.5K]2 years ago
4 0

Answer:

Hey

I have no idea when YOUR assignment is due.

Newtons 1rst law:

An object that has constant motion will remain at that speed unless acted on by an external force.

Newtons 2nd law:

F=ma (force=mass*acceleration)

Newtons 3rd law:

when a force is applied to an object, there will be an opposite but equal reaction.

Acceleration:

How much your speed increases/decreases per unit of time.

I wrote all that^

You might be interested in
Negative charges can move from one insulator to another.<br> True<br> False
Rasek [7]

Answer:

True

Explanation:

Because in atom the negative charge become lose or gain.

6 0
3 years ago
PLEASEEEEEE HEEEEEEEEEEEEEEELPPPPPPPPPPPPPPPPP I NED HELP WHIT THIS!!!!! D:
aleksley [76]
The meters per second
+1t a second / 2t
6 0
2 years ago
Find the acceleration that can result from a net force of 13 N exerted on a 3.6-kg cart. (Note: The unit N/kg is equivalent to m
STatiana [176]

Answer:

a = 3.61[m/s^2]

Explanation:

To find this acceleration we must remember newton's second law which tells us that the total sum of forces is equal to the product of mass by acceleration.

In this case we have:

F = m*a\\\\m=mass = 3.6[kg]\\F = force = 13[N]\\13 = 3.6*a\\a = 3.61[m/s^2]

7 0
3 years ago
A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?
stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, m_1 = 300\ g = 0.3\ kg

Initial speed of the cart, u_1=1.2\ m/s

Mass of the larger cart, m_2 = 2\ kg

Initial speed of the larger cart, u_2=0

After the collision,

Final speed of the smaller cart, v_1=-0.81\ m/s (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let v_2 is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

m_1u_1+m_2u_2-m_1v_1=m_2v_2

v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}

v_2=0.301\ m/s

So, the speed of the large cart after collision is 0.301 m/s.

4 0
3 years ago
you stop the stopwatch at 4.0 s, but you notice a short time later that the same ant is at 0.81 m on the meter stick. Assuming t
telo118 [61]

The time elapsed since you stopped the stopwatch is 0.41 s.

<em>Your question is not complete, it seems to be missing the following information;</em>

"The velocity of the ant is 2 m/s"

The given parameters;

  • velocity of the ant, v = 2 m/s
  • change in position of the ant, Δx = 0.81 m
  • initial time, t₁ = 4 s
  • time when the ant was noticed, = t₂

Velocity is defined as the change in displacement per change in time of motion of an object.

v = \frac{\Delta x}{\Delta t} = \frac{\Delta x}{t_2 - t_1} \\\\t_2 -t_1 = \frac{\Delta x}{v} \\\\t_2 - 4 = \frac{0.81}{2} \\\\t_2 - 4 = 0.405\\\\t_2 = 0.405 + 4\\\\t_2 = 4.405 \approx 4.41 \ s

The time elapsed since you stopped the stopwatch is calculated as;

t_{elapsed} = 4.41 \ s - 4\ s = 0.41 \ s

Thus, the time elapsed since you stopped the stopwatch is 0.41 s.

Learn more here: brainly.com/question/18153640

5 0
3 years ago
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