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3 years ago

Help please (about projectile motion at an angle)

Physics

1 answer:

PolarNik [594]3 years ago

7 0

Answer:

Since this is old, im just gonna get these points, don't wan't them to go to waste lm.ao

Explanation:

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What is the current if 4C of charge passes in 2 s?

julia-pushkina [17]

Answer:

I hope 2 amperes of current passes

8 0

2 years ago

Intelligence tests that are given to participants in groups

White raven [17]

a. are typically paper-and-pencil measures.

Similar with psychological tests, mostly structured personality tests.

Psychological tests comes two ways: 
The structure psychological tests or, objectives tests and unstructured psychological tests or, also called projective tests. By what you are referring the responder strongly asserts a projective tests which in definition comes with an unambiguous stimuli or no paper test just drawings and images. If what the responder’s suggesting is correct you are referring to the Rorschach projective tests, these tests are a figure symmetrically placed in an inkblot that lets you visualize or create a mental picture out of it, and makes you describe what you in see much detail as you can.

4 0

2 years ago

A 3.91 kg cart is moving at 5.7 m/s when it collides with a 4 kg cart which was at rest. They collide and stick together.

Nesterboy [21]

Answer:

The velocity after the collision is 2.82 m/s

Explanation:

Law Of Conservation Of Linear Momentum

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is

P=mv.

If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

$P=m_1v_1+m_2v_2$

If a collision occurs and the velocities change to v', the final momentum is:

$P'=m_1v'_1+m_2v'_2$

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

If both masses stick together after the collision at a common speed v', then:

$m_1v_1+m_2v_2=(m_1+m_2)v'$

The common velocity after this situation is:

$\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.

After the collision, both cars stick together. Let's compute the common speed after that:

$\displaystyle v'=\frac{3.91*5.7+4*0}{3.91+4}$

$\displaystyle v'=\frac{22.287}{7.91}$

$\boxed{v' = 2.82\ m/s}$

The velocity after the collision is 2.82 m/s

6 0

2 years ago

A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same di

Shalnov [3]

A distance of d is covered with 53 mile/hr initially. Time taken to cover this distance t1 = d/53 hour Next distance of d is covered with x mile hours. Time taken to cover this distance t2 = d/x hours. We have average speed = 26.5 mile / hour

= Total distance traveled/ total time taken

= $\frac{2d}{\frac{d}{53}+\frac{d}{x}} = \frac{2}{\frac{1}{53}+\frac{1}{x} } = \frac{106x}{x+53}$