All categories

3 years ago

A car moving with a speed of 35 m/s sees a child standing in the

Physics

1 answer:

zlopas [31]3 years ago

6 0

Answer:

2100 N

Explanation:

$v = u + at \\ 0 = 35 + a(5) \\ a = - 7m {s}^{ - 2} \\ \\ net \: force \: = ma \\ = (300)(7) \\ = 2100 \: newtons$

You might be interested in

a racing car undergoes a uniform acceleration of 4.00m/s2. if the net force causing the acceleration is 3.00 times 10^3 N, what

yKpoI14uk [10]

Answer: 750Kg

Explanation:

Recall that force is the product of the mass M, of an object moving at a uniform acceleration.

i.e Force = Mass x Acceleration

In this case, Mass = ?

Force = 3.00 x 10^3 N = (3.00 x 1000N)

= 3000N

Uniform acceleration = 4.00m/s^2

Force = Mass x Acceleration

3000N = Mass x 4.00m/s^2

Mass = (3000N/4.00m/s^2)

Mass = 750Kg (The SI unit of mass is kilograms)

Thus, the mass of the car is 750Kg

4 0

3 years ago

A 4.8 mF capacitor in series with a 500 Ω resistor is connected, by a switch, to a 12 V battery. The current through the resisto

ad-work [718]

Answer:

Current will be 81.7 mA

Which is not given in bellow option

Explanation:

We have given capacitance $C=4.8mF=4.8\times 10^{-3}F$

Resistance R = 500 ohm

Voltage V = 12 volt

We know that time constant of RC circuit of RC circuit is given by

$\tau =RC=500\times 4.8\times 10^{-3}=2.4sec$

Time is given as t = 1 sec

We know that current in RC circuit is given by

$i=\frac{v}{R}(1-e^{\frac{-t}{\tau }})$

So current $i=\frac{12}{500}(1-e^{\frac{-1}{2.4 }})=0.00817A=81.7mA$

Which is not given in the following option

3 0

3 years ago

Read 2 more answers

20 POINTSSS!!!!!!!!

Soloha48 [4]

Answer: A
Hope this help you!!

7 0

3 years ago

Unbalanced contact forces between several object would result in ?

Svetlanka [38]

In a reaction with a directing vendor of the sum forces towards some object or nothing

3 0

3 years ago

4. A 1200 kg car traveling North at 20.0 m/s collides with a 1400 kg car traveling South at 22.0 m/s. The two

Dvinal [7]

Answer:-2.61 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=mV_{o}+MU_{o}$ (2)

$p_{f}=(m+M)V_{f}$ (3)

$m=1200 kg$ is the mass of the first car

$V_{o}=20 m/s$ is the velocity of the first car, to the North

$M=1400 kg$ is the mass of the second car

$U_{o}=-22 m/s$ is the mass of the second car, to the South

$V_{f}$ is the final velocity of both cars after the collision

$mV_{o}+MU_{o}=(m+M)V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{mV_{o}+MU_{o}}{m+M}$ (5)

$V_{f}=\frac{(1200 kg)(20 m/s)+(1400 kg)(-22 m/s)}{1200 kg+1400 kg}$ (6)

Finally:

$V_{f}=-2.61 m/s$ (7) This is the resulting velocity of the wreckage, to the south

7 0

3 years ago