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Dmitry [639]
3 years ago
10

3. The car's mass is 400 kg. It moves at a velocity of 20 m/s. Calculate the car's momentum. *

Physics
1 answer:
Orlov [11]3 years ago
7 0

Answer:

momentum=mass×velocity

momentum =400kg×20m/s=8000kg.m/s

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Which measurement is a potential difference?<br> A.115J<br> B.115V<br> C.115N<br> D.115C
borishaifa [10]

Answer:

B

Explanation:

Potential difference has a SI Unit of Volt and its symbol is <em>V</em>. Hence answer is <u>B</u>.

A is wrong as it has the unit Joule <em>(J)</em> which is the SI unit for energy.

C is wrong as it has the unit Newton <em>(N)</em> which is the SI unit for force.

D is wrong as it has the unit Coulomb <em>(C)</em> which is the SI unit of charge.

5 0
4 years ago
Objects have a tendency to resist changing their motion. This property is called: *
Salsk061 [2.6K]

Answer:

The answer your looking for is option 2 - Inertia

3 0
3 years ago
A bowling sphere is sitting on the ground. One of your classmates
ikadub [295]

normal force because it is perpendicular to the surface

4 0
3 years ago
A mass on a spring A oscillates at twice the frequency of the same mass on spring B. Which statement is correct?A.The spring con
Nataliya [291]

Answer:

A.The spring constant for B is one quarter of the spring constant for A.

Explanation:

If spring A oscillates at twice the frequency of spring B, and period is frequency inverted. It means spring B has a period twice of spring A's.

T_B = 2T_A

As T = 2\pi\sqrt{\frac{m}{k}}, and the 2 springs have the same mass

2\pi\sqrt{\frac{m}{k_B}} = 2\pi\sqrt{\frac{m}{k_A}}

\sqrt{k_A} = 2\sqrt{B}

k_A = 4k_B

k_B = k_A/4

So A.The spring constant for B is one quarter of the spring constant for A. is the correct answer.

3 0
4 years ago
Consider a 150 turn square loop of wire 17.5 cm on a side that carries a 42 A current in a 1.7 T. a) What is the maximum torque
ankoles [38]

Answer:

(a) 328 Nm

(b) 79.35 Nm

Explanation:

N = =150, side = 17.5 cm = 0.175 m, i = 42 A, B = 1.7 T

A = side^2 = 0.175^2 = 0.030625 m^2

(a) Torque = N x i x A x B x Sinθ

For maximum torque, θ = 90 degree

Torque = 150 x 42 x 0.030625 x 1.7 x Sin 90

Torque = 328 Nm

(b) θ = 14 degree

Torque =  150 x 42 x 0.030625 x 1.7 x Sin 14

Torque = 79.35 Nm

7 0
3 years ago
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