All categories

3 years ago

How much work is done when a 1-kg mass is raised a vertical distance of 1 m

1 answer:

5 0

Weight = mass times gravity. . . . Work = (weight) x (height) = (mass) x (gravity) x (height) = (1) x (9.8) x (1) = 9.8 joules.

You might be interested in

What change if state occurs when water vapor from a hot shower forms small droplets of water on a bathroom mirror?

musickatia [10]

I think it becomes into water?

7 0

3 years ago

Read 2 more answers

Torque can be calculated by multiplying the force

lana66690 [7]

<span>Torque can be calculated by multiplying the force and the "perpendicular distance"

Hope this helps!</span>

6 0

3 years ago

What is the wavelength of an earthquake wave if it has a speed of 4 km/s and a frequency of 7 Hz?

damaskus [11]

Answer: 0.571 km/s

λ = v/f

λ = 4/7

λ = 0.571 km/s

7 0

4 years ago

Inside a 30.2 cm internal diameter stainless steel pan on a gas stove water is being boiled at 1 atm pressure. If the water leve

dybincka [34]

Answer:

Q = 20.22 x 10³ W = 20.22 KW

Explanation:

First we need to find the volume of water dropped.

Volume = V = πr²h

where,

r = radius of pan = 30.2 cm/2 = 15.1 cm = 0.151 m

h = height drop = 1.45 cm = 0.0145 m

Therefore,

V = π(0.151 m)²(0.0145 m)

V = 1.038 x 10⁻³ m³

Now, we find the mass of the water that is vaporized.

m = ρV

where,

m = mass = ?

ρ = density of water = 1000 kg/m³

Therefore,

m = (1000 kg/m³)(1.038 x 10⁻³ m³)

m = 1.038 kg

Now, we calculate the heat required to vaporize this amount of water.

q = mH

where,

H = Heat of vaporization of water = 22.6 x 10⁵ J/kg

Therefore,

q = (1.038 kg)(22.6 x 10⁵ J/kg)

q = 23.46 x 10⁵ J

Now, for the rate of heat transfer:

Rate of Heat Transfer = Q = q/t

where,

t = time = (18.6 min)(60 s/1 min) = 1116 s

Therefore,

Q = (23.46 x 10⁵ J)/1116 s

8 0

3 years ago

600 kg elephant runs at 5 m/s and jumpts onto a 100kg cart. If the coefficient of friction is .04 how far will the cart travel o

Minchanka [31]

Answer:

Eleven seconds.

Explanation:

Two keys are needed to solve this problem. First, the conservation of momentum: allowing you to calculate the cart's speed after the elephant jumped onto it. It holds that:

$m_ev_e+m_c\cdot0=(m_e+m_c)v_0\implies \\v_0=\frac{m_ev_e}{m_e+m_c}=\frac{600kg\cdot 5\frac{m}{s}}{700kg}=4.29\frac{m}{s}$

So, once loaded with an elephant, the cart was moving with a speed of 4.29m/s.

The second key is the kinematic equation for accelerated motion. There is one force acting on the cart, namely friction. The friction acts in the opposite direction to the horizontal direction of the velocity v0, its magnitude and the corresponding deceleration are:

$F_r = 0.04\cdot (m_e+m_c)g\implies a_r = 0.04\cdot g= 0.04 \cdot 9.8 \frac{m}{s^2}=0.39\frac{m}{s^2}$

The kinematic equation describing the decelerated motion is:

$v = -a_r t+v_0\\0 = -a_rt+v_0\implies \\t = \frac{v_0}{a_r}=\frac{4.29\frac{m}{s}}{0.39\frac{m}{s^2}}=11s$

It takes 11 seconds for the comical elephant-cart system to come to a halt.

7 0

4 years ago