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Digiron [165]
3 years ago
12

A 5 kg object is launched from a 35 m high cliff at an angle of 30 degrees with a speed of 100 m/s. How far from the cliff does

it hit the ground
Physics
1 answer:
yawa3891 [41]3 years ago
3 0

Answer:

The object will hit the ground at 882.77 m from the lunch

Explanation:

we are to find the range of the motion

What is range?

It is the distance from the point of projection to the point where the object hits the ground

<em>R=(v^2sin2θ)/g </em>

Given data

v= 100 m/s

θ= 30 degree

g= 9.81 m/s^2

Substituting to find the R we have

R= 100^2 *sin(2*30)/9.81

R= (10000*0.866)9.81

R= 8660/9.81

R= 882.77 m

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julsineya [31]

We will have that the diagram will be the following:

Here "F" is the force the shuttle is using to take off, friction is the friction with the air and "W" is the weight of the shuttle.

4 0
1 year ago
I’ll give brainliest, help!!
antiseptic1488 [7]

Answer:

A,W = 40 J

Explanation:

A, W =FS

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8 0
2 years ago
An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal butopposite charge on its plat
Basile [38]

Answer:

U_f = (U_o)/2)

Explanation:

The capacitance of a given capacitor is given by the formula;

C = ε_o•A/d

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Now,we are told that that all the dimensions of the capacitor plate is doubled. Thus, we now have;

C' = ε_o•4A/2d

Hence, C' = 2C

so capacitance is now doubled

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U_f = Q²/2C'

From earlier, we saw that C' = 2C.

Thus;

U_f = Q²/2(2C)

U_f = Q²/4C

Rearranging, we have;

U_f = (1/2)(Q²/2C)

From earlier, U_o = Q²/2C

Hence,

U_f = (1/2)(U_o)

Or

U_f = (U_o/2)

4 0
3 years ago
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anyanavicka [17]

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7 0
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The block A and attached rod have a combined mass of 50 kg and are confined to move along the guide under the action of the 796-
Luba_88 [7]

Answer:

The bending moment is 459.16 N.m

Explanation:

From the given information;

Let's assume that the angle is 66°

Then, the free body diagram is draw and attached in the file below.

Now, the calculation of the acceleration from the first part of the free body diagram is:

\sum F_x = ma_x \\ \\ 796 - 50(9.81) sin 66=50a \\ \\ 796 - 448.094 = 50 a  \\ \\ a = \dfrac{347.906}{50} \\ \\ a  = 6.96 \ m/s^2

Bending moment M:

From the second part of the diagram:

\sum M_B = mad \\ \\ M - (15 \times 9.81) (1.5) = (25 \times 6.96)(1.5 sin 66) \\ \\ M - 220.725 = 238.435  \\ \\  M = 238.435 + 220.725 \\ \\  \mathbf{M = 459.16 \ N.m}

6 0
3 years ago
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