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3 years ago

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A ball rolls off the top of the roof of a building that is 13 meters tall. Calculate the amount of time it takes for it to hit t

he ground.

1 answer:

kramer3 years ago

5 0

Answer:

The ball takes $t=1 \cdot 917sec$ to hit the ground.

Explanation:

• The second equation of motion represents the total distances travelled by an object in a time interval of $\Delta t$ with an initial speed of $u$ and acceleration $a$ .

• To find the time, ball takes to hit the ground, use the formula: $s = ut + \frac{1}{2}a{t^2}$

Where, $t$ is time, $u$ is initial velocity, $a$ is acceleration and $s$ is displacement.

• In this case, $a = g = 9 \cdot 8m/se{c^2}$ .

• Placing the value of the given initial velocity, $u=0cm/s$ and displacement, $s = 13m$ in the above formula.

$& \therefore s = ut + \frac{1}{2}g{t^2} \\& \Rightarrow 13 = 0 \cdot t + \frac{1}{2} \times 9 \cdot 8 \times {t^2} \\$

$& \Rightarrow 13 = 4 \cdot 9{t^2} \\& \Rightarrow {t^2} = \frac{{13}}{{4.9}} \\& \Rightarrow t = 1 \cdot 917sec \\\end{align}\]$

• Hence, ball takes $t=1 \cdot 917sec$ to hit the ground.

Learn more about second equation of motion here:

brainly.com/question/19030143

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Answer:u=42.29 m/s

Explanation:

Given

Horizontal distance=167 m

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Let u be the initial speed of ball

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The dome of a Van de Graaff generator receives a charge of 0.00011 C. The radius of the dome is 5.2 m. Find the strength of the

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Answer:

Answer:

Explanation:

Given that

K=8.98755×10^9Nm²/C²

Q=0.00011C

Radius of the sphere = 5.2m

g=9.8m/s²

1. The electric field inside a conductor is zero

εΦ=qenc

εEA=qenc

net charge qenc is the algebraic sum of all the enclosed positive and negative charges, and it can be positive, negative, or zero

This surface encloses no charge, and thus qenc=0. Gauss’ law.

Since it is inside the conductor

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2. Since the entire charge us inside the surface, then the electric field at a distance r (5.2m) away form the surface is given as

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The electric field at the surface of the conductor is 36561N/C

Since the charge is positive the it is outward field

3. Given that a test charge is at 12.6m away,

Then Electric field is given as,

E=kQ/r²

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3 years ago

The cable of the 1800kg elevator cab in Fig. 8−56 snaps when the cab is at rest at the first floor, where the cab bottom is a di

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a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

a ) The speed of the cab just before it hits the spring,

Ki + Pi = K final + P final + W

1 / 2 m vo² + m g hi = 1 / 2 m v² + m g h + f d

0 + ( 1800 * 9.8 * 3.7 m ) = ( 1 / 2 * 1800 * v² ) + 0 + ( 4400 * 3.7 )

v = 7.4 m / s

b ) The maximum distance x that the spring is compressed,

Ki + Pi = K final + P final + W + Fs

1 / 2 m vo² + m g x = 1 / 2 m v² + m g h + f d + 1 /2 k x²

( 1/2 * 1800 * 7.4² ) + ( 1800 * 9.8 * x ) =0 + 0+ ( 4400 * x ) + ( 1/2 * 1800 * x² )

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c ) The distance that the cab will bounce back up the shaft,

Ki + Pi + Fs = K final + P final + W

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h = 2.8 m

Therefore,

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

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A gray kangaroo can bound across level ground with each jump carrying it 9.1 m from the takeoff point. Typically the kangaroo le

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Explanation:

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