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Makovka662 [10]

2 years ago

8

A 30 kg box is placed on a skateboard and pushed across the room with a velocity of 4 m/s. The pusher applies more force, increa

sing the velocity to 10 m/s. How much work was done to move the box?

1 answer:

Reptile [31]2 years ago

3 0

Answer:

Dawg

Explanation:

Oh yea actually u suck

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Determine the magnitude of the momentum of a ... 107-kg halfback moving eastward at 8 m/s.The halfback's momentum in kgm/s is:

Veronika [31]

Given:

The mass of the halfback is m = 107 kg

The speed of the halfback is v = 8 m/s

To find the momentum.

Explanation:

The momentum of the halfback is

$\begin{gathered} p=\text{ mv} \\ =\text{ 107}\times8 \\ =\text{ 856 kg m/s} \end{gathered}$

Thus, the momentum of the halfback is 856 kg m/s

3 0

1 year ago

Si un cuerpo adquiere una carga de -0,02 C, ha ganado o ha perdido electrones? Cuantos?

sammy [17]

Question :

If a body acquires a charge of -0.02 C, has it gained or lost electrons? Many?

Solution :

We know, charge gained is shown by negative sign.

Since, charged acquired is given as -0.02 C .

Therefore, it is body has gained electrons.

Now, number of electrons is given by :

$n = \dfrac{net\ charge}{charge \ on \ one \ electron}\\\\n = \dfrac{-0.02}{-1.60 \times 10^{-19}}\\\\n = 1.25\times 10^{17}$

Hence, this is the required solution.

7 0

3 years ago

Which sentence in the passage describes retrograde motion of a planet

Soloha48 [4]

Answer:

what is the passage? I don’t see it

Explanation:

4 0

3 years ago

You are camping with two friends, Joe and Karl. Since all three of you like your privacy, you don't pitch your tents close toget

aleksklad [387]

Answer:

35.7 m

Explanation:

Let

$\mid A\mid=18.5 m$

$\mid B\mid=41 m$

We have to find the distance between Joe's and Karl'e tent.

$A_x=Acos\theta$

$A_y=Asin\theta$

Substitute the values then we get

$A_x=18.5cos23^{\circ}=17 m$

$A_y=18.5sin 23^{\circ}=7.2 m$

$B_x=41cos37.5^{\circ}=32.5 m$

$B_y=41sin37.5^{\circ}=-24.96 m$

Because vertical component of B lie in IV quadrant and y-inIV quadrant is negative.

By triangle addition of vector

$B=A+C$

$C=B-A$

$C_x=B_x-A_x=32.5-17=15.5 m$

$C_y=B_y-A_y=-24.96-7.2=-32.16\approx=-32.2 m$

$\mid C\mid=\sqrt{C^2_x+C^2_y}$

$\mid C\mid=\sqrt{(15.5)^2+(-32.2)^2}=35.7 m$

Hence, the distance between Joe's and Karl's tent=35.7 m

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2 years ago

You travel 85 miles in 2 hours, you then travel 75 miles in 1.5 hours. After a 0.5 hour rest break you then travel 55 miles in 1

Lera25 [3.4K]

Total 215 miles in 5 hours .hope it helps

8 0

3 years ago