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Zinaida [17]
3 years ago
15

A drag racer starts from the rest and accelerates at 7.4m/s^2.how far will he travel in 2.0 seconds?

Physics
1 answer:
Vesna [10]3 years ago
8 0

Answer:

14.8 m

Explanation:

S= ut + \frac{1}{2}at^{2}

where u = initial velocity

S= (0 \frac{m}{s})(2s) + \frac{1}{2}(7.4\frac{m}{s^{2} })(2s^{2})

S=  \frac{1}{2}(7.4\frac{m}{s^{2} })(2s^{2})

S=14.8 m

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At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f
jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp =  change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s,  0.65 kg m/s)

The magnitude of this vector is calculated as follows:

|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

4 0
3 years ago
It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the sp
Ede4ka [16]

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J

So, the new work is more than 130 J.

6 0
2 years ago
Please check my answers!
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B.) Areas near the equator have warmer climates than areas for form the equator.

D.) Winds that blow inland from oceans or large lakes contain a lot of water vapor that will cause precipitation.

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The magnitude of Alioth ( the brightest star in the big dipper ) is 1.76 and it is about 81 light years distant from Earth. 
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A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c
tensa zangetsu [6.8K]

Answer:

E = \frac{3kQ^2}{5R}

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

V = \frac{kq}{r}

now we can say that

q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)

q = \frac{Qr^3}{R^3}

now electric potential is given as

V = \frac{k\frac{Qr^3}{R^3}}{r}

V = \frac{kQr^2}{R^3}

now work done to bring a small charge from infinite to the surface of this sphere is given as

dW = V dq

dW = \frac{kQr^2}{R^3} dq

here we know that

dq = \frac{3Qr^2dr}{R^3}

now the total energy of the sphere is given as

E = \int dW

E = \int_0^R  \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})

E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)

E = \frac{3kQ^2}{5R}

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