When work is done, an object changes its

The kettle is filled with 5.0 kg of water at 20 °C. The specific heat capacity of water = 4 200 J/Kg °C. Calculate the energy ne

Allisa [31]

Answer:

17 * 10^5 J

Explanation:

From;

H= mcΘ

where;

m = mass of water = 5.0 Kg

c = heat capacity of water = 4 200 J/Kg °C

Θ = temperature rise = 100 °C - 20 °C = 80°C

Substituting values;

H = 5 * 4 200 * 80

H = 17 * 10^5 J

4 0

3 years ago

There is a bell at the top of a tower that is 95 m high. The mass of the bell is 21 kg. What is the potential energy of the bell

melomori [17]

Answer:

19,551 J!

Explanation:

The formula is PE = ham (h=height, a= acceleration or 9.8, m= mass)

PE = (95)(9.8)(21)

PE = 19,551 Joules

5 0

3 years ago

A 60 kg swimmer at a water park enters a pool using a 2 m high slide. Find the velocity of the swimmer

koban [17]

Answer:

the velocity of the swimmer at the bottom of the slide is 6.26 m/s

Explanation:

The computation of the velocity of the swimmer at the bottom of the slide is given below:

v = √2gh

= √2 × 9.8 × 2

= 6.26 m/s

Hence, the velocity of the swimmer at the bottom of the slide is 6.26 m/s

5 0

3 years ago

Different forces were applied to five balls, and each force was applied for the same amount of time. The data is in the table. A

Annette [7]

The magnitude of impulse on Ball A will become twice (doubles). Hence, option (c) is correct.

Impulse:

The effect of applied force acting for a very small interval of time is known as Impulse.

Given data:

The initial magnitude of the force on Ball A is, F = 50 N.

The time interval for the applied force on Ball A is, t = 1.25 s.

And the impulse at initial on ball A is, I = 62.5 N-s.

Now, if force on ball A doubles, which means new magnitude of force is,

F' = 2F

F' = 2 × 50

F' = 100 N

For same time interval, the new impulse on Ball A is given as,

I' = F' × t

Solving as,

I' = 100 × 1.25

I' = 125 N-s

Taking the ratio of both the impulses as,

I' / I = 125/62.5

I' = 2 I

Thus, we can conclude that the magnitude of impulse on Ball A will become twice (doubles). Hence, option (c) is correct.

Learn more about the impulse here:

brainly.com/question/904448

7 0

3 years ago

An α-particle has a charge of +2e and a mass of 6.64 × 10-27 kg. It is accelerated from rest through a potential difference that

zzz [600]

Answer with Explanation:

We are given that

Charge on alpha particle=q=2 e= $2\times 1.6\times 10^{-19} C$

1 e= $1.6\times 10^{-19} C$

Mass of alpha particle=m= $6.64\times 10^{-27} kg$

Potential difference,V= $1.97\times 10^6 V$

Magnetic field,B=3.49 T

a.Speed of alpha particle=v= $\sqrt{\frac{2 qV}{m}}$

By using the formula

$v=\sqrt{\frac{2\times 2\times 1.6\times 10^{-19}\times 1.97\times 10^6}{6.64\times 10^{-27}}$

$v=1.38\times 10^7 m/s$

b.Magnetic force,F $=qvB=2\times 1.6\times 10^{-19}\times 1.38\times 10^7\times 3.49=1.5\times 10^{-11} N$

F= $1.5\times 10^{-11} N$

c.Radius of circular path, r= $\frac{mv^2}{F}$

$r= \frac{6.64\times 10^{-27}\times (1.38\times 10^7)^2}{1.5\times 10^{-11}}$

r=0.084 m

5 0

3 years ago