Ask question

Ask question

All categories

3 years ago

15

a given box of 2 kg is in contact with the floor experiences an applied force of 150 form somebody pushing it.while in motion,it

also experiences a frictional force of 7n in the opposite direction. Calculate the weight of the box

1 answer:

ladessa [460]3 years ago

6 0

Weight = (mass) x (gravity)

Weight = (2kg) x (9.8 m/s^2)

Weight = 19.6 N

You might be interested in

A. How many calories are needed to raise the temperature of 1 gram of water by 1 °C?

sweet-ann [11.9K]

A) 1 cal

B) 80 cal

C) 540 cal

Explanation:

A)

The amount of heat energy needed to raise the temperature of a certain mass of a substance is given by

$Q=mC\Delta T$

where

m is the mass of the substance

C is the specific heat capacity

$\Delta T$ is the change in temperature

In this problem:

m = 1 g is the mass of water

$C=1 cal/g^{\circ}C$ is the specific heat capacity of water

$\Delta T=1^{\circ}C$ is the change in temperature

So, the heat needed is

$Q=(1)(1)(1)=1 cal$

B)

For a solid substance at its melting point, the amount of heat needed to melt completely the substance is given by

$Q=m\lambda_f$

where

m is the mass of the substance

$\lambda_f$ is the specific latent heat of fusion of the substance

In this problem:

- The ice is already at melting point, 0 °C

- Mass of the ice: $m=1g$

- Specific latent heat of fusion of ice: $\lambda_f=80 cal/g$

So, the heat needed is

$Q=(1)(80)=80 cal$

C)

For a liquid substance at its boiling point, the amount of heat needed to boil completely the substance is given by

$Q=m\lambda_v$

where

m is the mass of the substance

$\lambda_v$ is the specific latent heat of vaporization of the substance

In this problem:

- The water is already at boiling point, 100 °C

- Mass of the water: $m=1g$

- Specific latent heat of vaporization of water: $\lambda_v=540 cal/g$

So, the heat needed is

$Q=(1)(540)=540 cal$

5 0

3 years ago

Suppose the price of electricity is 12.75 cents per kWºh. What

adell [148]

Answer:

The difference between the cost of operating LED and incandescent bulb is $5.1

Explanation:

We are given the cost of electricity that is 12.75 cents per kWh. We want to find out the difference in the operating cost of an incandescent and LED bulb for a time period of 2,000 hours.

Since we are not given the rating of the incandescent bulb and LED bulb, we will assume their ratings.

For a light intensity of 250 Lumens;

The average rating of an LED bulb is approximately 5 Watts.

The average rating of an incandescent bulb is approximately 25 Watts.

Now lets find out the kWh of each bulb.

Energy = Power×Time

For LED bulb:

E = 5×2,000 = 10,000 Wh

Divide by 1000 to convert into kWh

E = 10,000/1000 = 10 kWh

Cost = 12.75×10 = 127.5 cents

Cost = $1.27

For Incandescent bulb:

E = 25×2,000 = 50,000 Wh

Divide by 1000 to convert into kWh

E = 50,000/1000 = 50 kWh

Cost = 12.75×50 = 637.5 cents

Cost = $6.37

Difference in Cost:

Difference = $6.37 - $1.27 = $5.1

Therefore, the difference between the cost of operating LED and incandescent bulb is $5.1.

5 0

3 years ago

What is the anwser to impossible quiz question 66

Vinvika [58]

The ! at the end of the word world. click that

4 0

3 years ago

Read 2 more answers

A 1200 kg safe is 2.0 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses

makvit [3.9K]

<span>the spring constant is 200kN/m.
</span>
The work done on the spring by the the free falling safe is equal to the kinetic energy of the spring:

W1 = $\frac{1}{2}$ k $x^{2}$

The work done on the safe by gravity is the potential energy of the spring:

W2 = (mass of safe) x g x (height)
= 1000kg x 9.8 m/s^2 x (2 + 0.54)m
=1000kg x 9.8 m/s^2 x (2.54)m
= 24,892 N-m

W2 = W1 =24892 N-m
⇒ $\frac{1}{2}$ k $(0.5)^{2}$ = 24892 N-m

Therefore, k = $\frac{(24892)*2}{0.5*0.5}$ = 199136 kN/m.

The spring constant is 199,136 N/m ≈ 200 kN/m

7 0

3 years ago

In the Skycoaster amusement park ride, riders are suspended from a tower by a long cable. A second cable then lifts them until t

Alborosie

Answer:

T1 cos 15 = T2 cos 26

T1 = T2 cos 26 / cos 15

T1 = 0.930 T2 ------------(1)

T1 sin 15 + T2 sin 26 = m g

substitute eqn(1)

(0.930 T2 ) sin 15 + T2 sin 26 = 230 * 9.8

0.241 T2 + 0.438 T2 = 2254

Tension in right cable T2 = 3319 N

substitute in eqn (1)

T1 = 0.930 * 3319

Tension in left cable T1 = 3087

4 0

2 years ago

Read 2 more answers