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elena-14-01-66 [18.8K]
3 years ago
8

Please help me and thank you

Physics
1 answer:
fredd [130]3 years ago
8 0

Answer:

536.56 m/s

Explanation:

We'll begin by calculating the momentum of the Porsche. This can be obtained as follow:

Mass (m) of Porsche = 1361 kg

Velocity (v) of Porsche = 26.82 m/s

Momentum of Porsche =?

Momentum = mass × velocity

Momentum = 1361 × 26.82

Momentum of Porsche = 36502.02 Kgm/s

Finally, we shall determine the velocity you need to be running with in order to have the same momentum as the Porsche. This can be obtained as follow:

Your Mass = 68.03 kg

Your Momentum = Momentum of Porsche = 36502.02 Kgm/s

Your velocity =?

Momentum = mass × velocity

36502.02 = 68.03 × velocity

Divide both side by 68.03

Velocity = 36502.02 / 68.03

Velocity = 536.56 m/s

Thus you must be running with a speed of 536.56 m/s in order to have the same momentum as Porsche.

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According to Newton's second law, the resultant of the forces acting on the box is equal to the product between its mass and its acceleration:
\sum F = ma (1)

we are only concerned about the horizontal direction, so there are only two forces acting on the box in this direction:
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F_x = F cos \theta = (250N)(\cos 32.0^{\circ} )=212.0 N
- the frictional force, acting in the opposite direction, which is equal to
F_f = \mu mg=(0.350)(50.0 kg)(9.81 m/s^2)=171.7 N

By applying Newton's law (1), we can calculate the acceleration of the box:
F_x - F_f = ma
a= \frac{F_x - F_f}{m}= \frac{212.0 N-171.7 N}{50.0 kg} =0.81 m/s^2

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Why do waves in a tank are 10 cm long if they pass a point at a rate of 3.75 waves per second what is the speed in m/s?
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A 20 cm-radius ball is uniformly charged to 71 nC.
artcher [175]

Answer:

Part a)

\rho = 2.12\mu C/m^3

Part b)

q_1 = 1.11 nC

q_2 = 8.88 nC

q_3 = 71 nC

Part c)

E_1 = 3996 N/C

E_2 = 7992 N/C

E_3 = 15975 N/C

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

V = \frac{4}{3}\pi R^3

V = \frac{4}{3}\pi(0.20)^3

V = 0.0335 m^3

now the charge density of the ball is given as

\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3

Part b)

Now the charge enclosed by the surface is given as

q = \rho V

at radius of 5 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3

q = 1.11 nC

at radius of 10 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3

q = 8.88 nC

at radius of 20 cm

q = 71 nC

Part c)

As we know that electric field is given as

E = \frac{kq}{r^2}

so we have electric field at r = 5 cm

E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}

E_1 = 3996 N/C

electric field at r = 10 cm

E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}

E_2 = 7992 N/C

electric field at r = 20 cm

E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}

E_3 = 15975 N/C

3 0
3 years ago
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