Ask question

Ask question

All categories

3 years ago

13

The speed of a 2.0-kg object changes from 30 m/s to 40 m/s during a 5.0-second time interval.

1 answer:

AnnyKZ [126]3 years ago

5 0

Answer:

F = 2 * 30 / 5 = 12 N to stop forward motion

F = 2 * 40 / 5 = 16 N to accelerate to 90 degrees

(12^2 + 16^2)^1.2 = 20 N average force applied

You might be interested in

What is the upward force that balances the weight of an object on a surface

Alina [70]

Answer:

The support force/ Normal reaction force

Explanation:

Refer to Newtons third law.

"The third law states that when a force acts on a body due to another body, then an equal and opposite force acts simultaneously on that body"

Since we have a downward force acting on your object, in this case gravity/weight, there is an equal and opposite force given off by the surface the object is on. This is called the support force or the normal reaction force

7 0

3 years ago

Read 2 more answers

The base ( foundation ) of building is made wider. Why ?

attashe74 [19]

Answer: It is increased to exert the pressure off the bywalls and on large area (brainliest pls)

Explanation: ...

7 0

3 years ago

What is measured in watts that starts with a p

Nesterboy [21]

Power is measured in watts

6 0

3 years ago

A car accelerates at a rate of 3 m/s^2. If its original speed is 8 m/s, how many seconds will it take the car to reach a final s

kkurt [141]

Answer:

$\Large \boxed{\mathrm{5.67 \ seconds }}$

Explanation:

$\displaystyle \mathrm{acceleration \ = \ \frac{final \ velocity - initial \ velocity }{elapsed \ time}}$

$\displaystyle A \ = \ \frac{V_f - V_i }{t}$

$\displaystyle 3 \ = \ \frac{25 - 8 }{t}$

$\displaystyle 3 \ = \ \frac{17 }{t}$

$\displaystyle t \ = \ \frac{17 }{3} \approx 5.67$

4 0

3 years ago

How much heat must be removed from 456 g of water at 25.0°C to change it into ice at - 10.0°C ? The specific heat of ice is 2090

cupoosta [38]

Answer:209.98 kJ

Explanation:

mass of water $m=456 gm$

Initial Temperature of Water $T_i=25^{\circ}C$

Final Temperature of water $T_f=-10^{\circ}C$

specific heat of ice $c=2090 J/kg-K$

Latent heat $L=33.5\times 10^4 J/kg$

specific heat of water $c_{water}=4.184 KJ/kg-K$

Heat require to convert water at $T=25^{\circ}C$ to $T=0^{\circ}C$

$Q_1=0.456\times 4.184\times (25-0)=47.69 kJ$

Heat require to convert water at $T=0^{\circ}$ to ice at $T=0^{\circ}$

$Q_2=m\times L=0.456\times 33.5\times 10^4=152.76 kJ$

heat require to convert ice at $T=0^{\circ} C\ to\ T=-10^{\circ} C$

$Q_3=0.456\times 2090\times (0-(-10))=9.53 kJ$

Total heat $Q=Q_1+Q_2+Q_3$

$Q=47.69+152.76+9.53=209.98 kJ$

7 0

3 years ago