All categories

2 years ago

Acceleration of a refrigerator 30s after a person begins pushing it at a force of 400 N

1 answer:

6 0

Question: Find acceleration of a refrigerator 30s after a person begins pushing it at a force of 400 N, If the mass of the refrigerator is 10 kg.

Answer:

40 m/s²

Explanation:

Applying,

F = ma................Equation 1

Where F = Force applied to the refrigerator, m = mass of the refrigerator, a = acceleration of the refrigerator.

make a the subject of the equation

a = F/m............ Equation 2

From the question,

Given: F = 400 N, m = 10 kg

Substitute these values into equation 2

a = 400/10

a = 40 m/s²

You might be interested in

A 6 kg object falls 10 m. The object is attached mechanically to a paddle-wheel which rotates as the object falls. The paddle-wh

maksim [4K]

Answer: $16.234^{\circ}C$

Explanation:

Given

Mass of object (m)=6 kg

falling height(h)=10 m

mass of water( $m_w$ )=600 gm

temperature of water =15

specific heat of water $=4.186 j/g-^{\circ}C$

Let T be the Final Temperature of water

Here Object Potential Energy is converted into Heat energy which will be absorbed by water

Potential Energy(P.E.) $=mgh=6\times 9.81\times 10=588.6 J$

Heat supplied $=m_wc(\Delta T)$

$H.E.=600\times 4.186\times (T-16)$

$588.6=2511.6\times (T-16)$

T-16=0.234

$T=16.234^{\circ}C$

This is not an efficient way of heating water as there is only $0.234^{\circ}C$ increase in temperature.

5 0

3 years ago

Followed by the previous question: presume that the electron performs a uniform circular motion around the hydrogen nucleus. Wha

Ksivusya [100]

Answer:

$A_c=87.73*10^{21}m/s$

Explanation:

From the question we are told that

$r=5\times 10^{-11}$

$T=1.5 \times 10^{-16}$

Generally the equation for velocity is mathematically given as

$Velocity (v)=\frac{2 \pi r}{t}$

$V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}$

Generally the equation for Centripetal acceleration is mathematically given as

$A_c=\frac{V^2}{r}$

$A_c=(\frac{20.944*10^5)}{r5*10^{-11}}$

$A_c=87.73*10^{21}m/s$

8 0

2 years ago

The two-second rule applies to any speed __________ A. in all weather conditions. B. in ideal weather and road conditions. C. in

kumpel [21]

Answer:

The answer is A

Explanation:

A large risk of tailgating is the collision avoidance time being much less than the driver reaction time. Driving instructors advocate that drivers always use the "two-second rule" regardless of speed or the type of road. During adverse weather, downhill slopes, or hazardous conditions such as black ice, it is important to maintain an even greater distance.

5 0

3 years ago

Read 2 more answers

11. Which of the following is a proper unit of

Sergeeva-Olga [200]

D is the correct answer!!

4 0

2 years ago

Read 2 more answers

a 2,000 pound car is driving at 60 miles/hour along a straight, level road. what is the net force acting on the car?

SVETLANKA909090 [29]

Answer: