What is the half-life of an isotope that decays to 25% of its original activity in 70.8 hours?

If 3600 j of work is done in 3.0 s what is the power<br>0.00083W<br>1200W<br>3600W<br>11000W

Viktor [21]

Answer:

1200 W

Explanation:

Power is given by the ratio between work done and time taken:

$P=\frac{W}{t}$

where W is the work done and t the time taken.

In this problem, W = 3600 J and t = 3.0 s. Therefore, the power in this exercise is

$P=\frac{3600 J}{3.0 s}=1200 W$

5 0

3 years ago

Read 2 more answers

How much force is required to accelerate a 50 kg mass at 2 m/s

marta [7]

Answer:

F = 100 Newtons

Explanation:

F = ma

F = (50 kg)(2 m/s)

F = 100 kg*m/s

3 0

2 years ago

You take a trip from your home to Mali, a very poor country in western Africa. How is the water usage you observe there likely t

Harlamova29_29 [7]

Answer:

It is likely that the citizens have very limited water for agriculture and personal uses and it is stored in their homes, while in the U.S we have water ready at the moment we need it and can access it with faucets.

Explanation:

4 0

2 years ago

A bowling ball of 35.2kg, generates 218 kg.m/s units of momentum. What is the velocity of the bowling ball?

KIM [24]

Answer:

6.19m/s

Explanation:

Given parameters:

Mass of bowling ball = 35.2kg

Momentum = 218kgm/s

Unknown:

Velocity of the bowling ball = ?

Solution:

Momentum is the quantity of motion a body possess;

Momentum = mass x velocity

218 = 35.2 x velocity

Velocity = 6.19m/s

7 0

3 years ago

A canister is filled with 310 g of ice and 100. G of liquid water, both at 0 ∘c . The canister is placed in an oven until all th

olya-2409 [2.1K]

Here in the process we require

1. Heat to melt down all ice

2. Heat to raise the temperature of whole water to 100 degree C

3. Heat to boil off the water

now here for the first part

Heat required to melt the ice

$Q_1 = mL$

$Q_1 = 310*80 = 24800 cal$

now heat required to raise the temperature to 100 degree C

$Q_2 = ms\Delta T$

$Q_2 = (310 + 100)*1*(100 - 0)$

$Q_2 = 41000 cal$

Now heat required to boil it off

$Q_3 = mL$

$Q_3 = 410*540 = 221400 cal$

now the total heat required will be

$Q = Q_1 + Q_2 + Q_3$

$Q = 24800 + 41000 + 221400$

$Q = 287200 cal$

so it required 287200 calorie heat to boil it all water

6 0

3 years ago