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3 years ago

1. What is the potential energy of a 5.0-kg

Physics

1 answer:

babymother [125]3 years ago

5 0

Answer:

C. 98 J

Explanation:

The appropriate formula is ...

PE = mgh . . . . . m is mass; below, m is meters

PE = (5 kg)(9.8 m/s^2)(2 m) = 98 kg·m^2/s^2

PE = 98 J

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a concrete slab 20 m long and weighing 400,000 N is supported by one pillar. A 19,600 N car is parked 8 meters from one end, whe

Elden [556K]

let the distance of pillar is "r" from one end of the slab

So here net torque must be balance with respect to pillar to be in balanced state

So here we will have

$Mg(r - L/2) = mg(L/2 - 8)$

here we know that

mg = 19600 N

Mg = 400,000 N

L = 20 m

from above equation we have

$400,000(r - 10) = 19,600 (10 - 8)$

$r - 10 = 0.098$

$r = 10.098 m$

so pillar is at distance 10.098 m from one end of the slab

3 0

3 years ago

A 26.2-kg dog is running northward at 3.02 m/s, while a 5.30-kg cat is running eastward at 2.74 m/s. Their 65.1-kg owner has the

REY [17]

Answer:

Angle with the +x axis is θ = 79.599degree

Then the velocity of owner = 1.235m/s

Explanation:

Given that the mass of dog is m1 =26.2 kg

velocity of dog is u1 = 3.02 m/s (north)

mass of cat is m2 = 5.3 kg

velocity is u2 = 2.74 m/s (east )

Mass of owner is M = 65.1 kg

Consider the east direction along +x axis andnorth along +y

momentum of dog is Py = m1 x u1

= 79.124 kg.m/s (j)

momentum of cat is Px = m2 x u2

= 14.522 kg.m/s (i)

Then the net magnitude of momentum is P = (Px2 + Py2)1/2

= 80.445

Angle with the +x axis is θ =tan-1(Py / Px ) = 79.599 degree

Then the velocity of owner is v = P / M = 1.235 m/s

3 0

3 years ago

An unstable atomic nucleus has a mass of 17.010-27kg, and starts out at rest. When it decays, it the original nucleus disintegra

slega [8]

Answer:

Part a)

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

$E = 4.4 \times 10^{-13} J$

Explanation:

As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same

So we will have

$m_1v_1 + m_2v_2 + m_3v_3 = 0$

$(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0$

$(30\hat j + 33.6\hat i)\times 10^6 + 3.6 v = 0$

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

By equation of kinetic energy we have

$E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2$

$E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}$

$E = 9\times 10^{-14} + 6.72 \times 10^{-14} + 2.82\times 10^{-13}$

$E = 4.4 \times 10^{-13} J$

8 0

3 years ago

Which of the following gases are the heaviest? <br> O2, CH4, CO2, Cl2

kvasek [131]

the answer is

CI2 because its 70.

5 0

3 years ago

Puck 1 (1 kg) travels with velocity 20 m/s to the right when it collides with puck 2 (1 kg) which is initially at rest. After th

Burka [1]

Answer:

Explanation:

Parameters given:

Mass of Puck 1, m = 1 kg

Mass of Puck 2, M = 1 kg

Initial velocity of Puck 1, u = 20 m/s

Initial velocity of Puck 2, U = 0 m/s

Final velocity of Puck 1, v = 5 m/s

Since we are told that momentum is conserved, we apply the principle of conservation of momentum:

Total initial momentum of the system = Total final momentum of the system

mu + MU = mv + MV

(1 * 20) + (1 * 0) = (1 * 5) + (1 * V)

20 = 5 + V

V = 20 - 5 = 15 m/s

Puck 2 moves with a velocity of 15 m/s

7 0

3 years ago