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2 years ago

Can someone please give me the (Answers) to this? ... please ... I need help….

Physics

1 answer:

tensa zangetsu [6.8K]2 years ago

6 0

Answer:

1320 kg I think

Explanation:

Here you can just substitute into the equation. ∆v means change in speed

this gives you (1100x24)/20

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if a stone is projected at an angle of 50 degrees to the horizontal with an initial velocity of 50m/s, what is the vertical comp

Vilka [71]

Answer:

38.3 m/s

Explanation:

To find vertical component of initial velocity, you'd have to use sine ratio:

$\displaystyle{\sin \theta = \dfrac{u_y}{u}}$

$\displaystyle{u_y}$ is vertical component of initial velocity and $\displaystyle{u}$ is initial velocity given which is 50 m/s.

A stone is projected at an angle of 50 degrees so $\displaystyle{\theta}$ = 50°. Substitute in the formula:

$\displaystyle{\sin 50^{\circ} = \dfrac{u_y}{50}}\\\\\displaystyle{50 \sin 50^{\circ} = u_y}\\\\\displaystyle{u_y = 38.3 \ \, \sf{m/s}}$

Therefore, the vertical component of initial velocity is approximately 38.3 m/s

(The picture is also attached for visual reference!)

3 0

2 years ago

A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand

enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

$v=gt$

And the distance traveled downwards is:

$\displaystyle y=\frac{gt^2}{2}$

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

$\displaystyle t=\sqrt{\frac{2y}{g}}$

Replacing into the first equation:

$\displaystyle v=g\sqrt{\frac{2y}{g}}$

Rationalizing:

$\displaystyle v=\sqrt{2gy}$

Let's call v1 the final speed of the package dropped from a height H. Thus:

$\displaystyle v_1=\sqrt{2gH}$

Let v2 be the final speed of the package dropped from a height 4H. Thus:

$\displaystyle v_2=\sqrt{2g(4H)}$

Taking out the square root of 4:

$\displaystyle v_2=2\sqrt{2gH}$

Dividing v2/v1 we can compare the final speeds:

$\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}$

Simplifying:

$\displaystyle v_2/v_1=2$

The final speed of the second package is twice as much as the final speed of the first package.

5 0

3 years ago

You have an initial velocity of -3.0 m/s. You then experience an acceleration of 2.5 m/s2 for 9.0s; what

Tatiana [17]

27.9 idkkkk look it up on photomath

8 0

3 years ago

The work done by an engine equals one-fourth the energy it absorbs from a reservoir. True or False

Lunna [17]

True IF the engine is 25% efficient. False otherwise.

7 0

3 years ago

A potter's wheel has the shape of a solid uniform disk of mass 7 kg and radius 0.65 m. It spins about an axis perpendicular to t

ozzi

Answer:

1.832 kgm^2

Explanation:

mass of potter's wheel, M = 7 kg

radius of wheel, R = 0.65 m

mass of clay, m = 2.1 kg

distance of clay from centre, r = 0.41 m

Moment of inertia = Moment of inertia of disc + moment f inertia of the clay

I = 1/2 MR^2 + mr^2

I = 0.5 x 7 x 0.65 x 0.65 + 2.1 x 0.41 x 0.41

I = 1.47875 + 0.353

I = 1.832 kgm^2

Thus, the moment of inertia is 1.832 kgm^2.

5 0

3 years ago

￼Can someone please give me the (Answers) to this? ... please ... I need help….

Can someone please give me the (Answers) to this? ... please ... I need help….