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lara31 [8.8K]
2 years ago
9

NEED HELP!!!!! 10 POINTS

Physics
2 answers:
castortr0y [4]2 years ago
6 0

Answer: D.

Explanation:

satela [25.4K]2 years ago
4 0
The answer is D I’m pretty sure
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A car starting from rest (i.e. initial velocity = 0.0 m/s), moves in the positive X-direction with a constant average accelerati
olga_2 [115]

Answer:

Final speed of the car, v = 24.49 m/s

Explanation:

It is given that,

Initial velocity of the car, u = 0

Acceleration, a=3.1\ m/s^2

Time taken, t = 7.9 s

We need to find the final velocity of the car. Let it is given by v. It can be calculated using first equation of motion as :

v=u+at

v=0+3.1\times 7.9

v = 24.49 m/s

So, the final speed of the car is 24.49 m/s. Hence, this is the required solution.

8 0
3 years ago
Một chất điểm khối lượng m=200g chuyển
MrMuchimi
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2 years ago
What is a satellite
g100num [7]

Answer:

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Explanation:

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7 0
3 years ago
Read 2 more answers
What mass of a material with density rho is required to make a hollow spherical shell having inner radius r1 and outer radius r2
Margarita [4]

Answer:

m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3  )

Explanation:

  • We have to make a hollow sphere of inner  radius r_1 and outer radius r_2.

Then the mass of the material required to make such a sphere would be calculated as:

Total volume of the spherical shell:

V_t=\frac{4}{3} \pi.r_2^3

And the volume of the hollow space in the sphere:

V_h=\frac{4}{3} \pi.r_1^3

Therefore the net volume of material required to make the sphere:

V=V_t-V_h

V=\frac{4}{3} \pi(r_2^3-r_1^3)

  • Now let the density of the of the material be \rho.

<u>Then the mass of the material used is:</u>

m=\rho.V

m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3  )

4 0
3 years ago
the total positive charge is QQQ = 1.62×10−6 CC , what is the magnitude of the electric field caused by this charge at point P,
balu736 [363]

Answer:

6.1 × 10^9 Nm-1

Explanation:

The electric field is given by

E= Kq/d^2

Where;

K= Coulombs constant = 9.0 × 10^9 C

q = magnitude of charge = 1.62×10−6 C

d = distance of separation = 1.53 mm = 1.55 × 10^-3 m

E= 9.0 × 10^9 × 1.62×10−6/(1.55 × 10^-3 )^2

E= 14.58 × 10^3/2.4 × 10^-6

E= 6.1 × 10^9 Nm-1

8 0
3 years ago
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