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2 years ago

An object with a mass of 5kg accelerates at 2m/s2. How much force in Newtons(N) is needed to cause this to happen?

Physics

1 answer:

Dmitry [639]2 years ago

4 0

Force = Mass X Acceleration

F = 5 x 2

F = 10 N

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(b) During one day, 250 kg of water is pumped through

ser-zykov [4K]

Answer: The energy incident on the solar panel during that day is $9.24 \times 10^{7} J$ .

Explanation:

Given: Mass = 250 kg

Initial temperature = $16^{o}C$

Final temperature = $38^{o}C$

Specific heat capacity = 4200 $J/kg^{o}C$

Formula used to calculate the energy is as follows.

$q = m \times C \times (T_{2} - T_{1})$

where,

q = heat energy

m = mass of substance

C = specific heat capacity

$T_{1}$ = initial temperature

$T_{2}$ = final temperature

Substitute the values into above formula as follows.

$q = 250 kg \times 4200 J/kg^{o}C \times (38 - 16)^{o}C\\= 250 kg \times 4200 J/kg^{o}C \times 22^{o}C$

As it is given that water absorbs 25% of the energy incident on the solar panel. Hence, energy incident on the solar panel can be calculated as follows.

$\frac{25}{100} \times q = 250 kg \times 4200 J/kg^{o}C \times 22^{o}C\\q = 9.24 \times 10^{7} J$

Thus, we can conclude that the energy incident on the solar panel during that day is $9.24 \times 10^{7} J$ .

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2 years ago

An object is at x = 0 at t = 0 and moves along the x axis according to the velocity–time graph in Figure P2.50.(a) What is the o

bezimeni [28]

Picture? I may be able to answer if you have a chart or some kind of graph as a referral to the question

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3 years ago

Which of the following best describes what went wrong with the scientists’ study? an improper experimental procedure the lack of

Rasek [7]

B- lack of control.

the scientists did not have a proper control group.

3 0

3 years ago

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A 15g bullet is fired horizontally into a 3kg block of wood suspended by a long cord. Assume that the bullet remains in the bloc

Varvara68 [4.7K]

Answer:

261.3 m/s

Explanation:

Mass of bullet=m=15 g= $\frac{15}{1000}=0.015 kg$

1 kg=1000g

Mass of block=M=3 kg

d=0.086 m

Total mass =M+m=3+0.015=3.015 kg

K.E at the time strike=Gravitational potential energy at the end of swing

$\frac{1}{2}(m+M)^2V^2=(m+M)gh$

Using g= $9.8m/s^2$

Substitute the values

$\frac{1}{2}(3.015)V^2=3.015\times 9.8\times 0.086$

$V^2=\frac{2\times 3.015\times 9.8\times 0.086}{3.015}$

$V=\sqrt{2\times 3.015\times 9.8\times 0.086}}$

$V=1.3m/s$

Velocity after collision=V=1.3 m/s

Velocity of block=v'=0

Using conservation law of momentum

$mv+Mv'=(m+M)V$

Using the formula

$0.015v+3(0)=3.015(1.3)$

$0.015v=3.015(1.3)$

$v^2=\frac{3.015(1.3)}{0.015}=261.3$

$v=261.3 m/s$

5 0

3 years ago

Pls help on this one

lidiya [134]

Answer:

Explanation:

because 50.0/10.0 = 5.0

5 0

2 years ago

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