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3 years ago

What is the potential energy of a 20-kg safe sitting on a shelf 0.5 meters

Physics

1 answer:

ki77a [65]3 years ago

8 0

Explanation:

P.E=MGH

Where m is mass

Where G is Acceleration Due to Gravity

Where h is Height

So the parameters are M = 20kg

G = 9.8m/s

H = 0.5meters

P.E= 20x9.8x0.5

=98J.

So Ans. is A= 98J

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Why do we balance chemical equations

antiseptic1488 [7]

So it could follow the correct mass for the atom

5 0

3 years ago

A watermelon is dropped from the top of a 80m tall building. We want to find the velocity of the watermelon after it falls for 1

White raven [17]

Answer:

$v_f = v_i + at$

$v_f = 13.23 m/s$

Explanation:

Height Of the watermelon when it is dropped is given as

$h = 80 m$

time of fall under gravity

$t = 1.35 s$

now if water melon start from rest then we have

$v_i = 0$

acceleration due to gravity for watermelon

$a = 9.81 m/s^2$

now we need to find the final speed of watermelon

$v_f = v_i + at$

so we will have

$v_f = 0 + (9.81)(1.35)$

$v_f = 13.23 m/s$

7 0

4 years ago

Two cars, one of mass 1400 kg, and the

Vikki [24]

Suppose that, in the x-y plane, the first car is moving to the right so that its velocity is given by the vector

v₁ = (14 m/s) i

and the second car is moving upward so that its velocity vector is

v₂ = (20 m/s) j

Then the total momentum of two cars before their collision is

m₁v₁ + m₂v₂ = (1400 kg) (14 m/s) i + (2300 kg) (20 m/s) j

= (19,600 i + 46,000 j) kg•m/s

Their momentum after the collision is

(1400 kg + 2300 kg) v = (3700 kg) v

where v is the velocity vector of the wreckage.

By conservation of momentum,

(19,600 i + 46,000 j) kg•m/s = (3700 kg) v

Let a and b be the horizontal and vertical components of v, respectively. Then

19,600 kg•m/s = (3700 kg) a ⇒ a ≈ 5.2973 m/s ≈ 5.3 m/s

46,000 kg•m/s = (3700 kg) b ⇒ b ≈ 12.4324 m/s ≈ 12 m/s

so that the final speed of the wreckage is

||v|| = √(a² + b²) ≈ 13.5139 m/s ≈ 14 m/s

3 0

2 years ago

A jet, sitting on the runway, takes off and accelerates at 8.0 m/s for 16s. How far did the jet travel down the runway?

nydimaria [60]

Answer:

2.4 m/s". 1

Explanation:

A jet with mass m = 8 x 10* kg jet accelerates down the runway for takeoff at 2.4 m/s". 1

8 0

3 years ago

A uniform stick 1.5 m long with a total mass of 250 g is pivoted at its center. A 3.3-g bullet is shot through the stick midway

Andru [333]

Answer:

63.44 rad/s

Explanation:

mass of bullet = 3.3 g = 0.0033 kg

initial velocity of bullet $v_{1}$ = 250 m/s

final velocity of bullet $v_{2}$ = 140 m/s

loss of kinetic energy of the bullet = $\frac{1}{2}m(v^{2} _{1} - v^{2} _{2})$

==> $\frac{1}{2}*0.0033*(250^{2} - 140^{2} )$ = 70.785 J

this energy is given to the stick

The stick has mass = 250 g =0.25 kg

its kinetic energy = 70.785 J

from

KE = $\frac{1}{2} mv^{2}$

70.785 = $\frac{1}{2}*0.25*v^{2}$

566.28 = $v^{2}$

$v= \sqrt{566.28}$ = 23.79 m/s

the stick is 1.5 m long

this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m

The angular speed will be

Ω = v/r = 23.79/0.375 = 63.44 rad/s

5 0

3 years ago