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3 years ago

15

Displacement vector of 4 km north, 2 km south, 5 km north, and 5km south combine to a total displacement of ? A 2 km South b 16

km south c 16 km north d 2km north?

1 answer:

Andrews [41]3 years ago

3 0

It is C. 16 km north.

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A 0.30-kg object is traveling to the right (in the positive direction) with a speed of 3.0 m/s. After a 0.20 s collision, the ob

andre [41]

Answer:

$I = -2.1 kg.m/s$

Explanation:

Given,

mass of the object,m = 0.30 Kg

initial speed, v_i = 3 m/s

time of collision = 0.20 s

final speed, v_f = -4 m/s

Impulse = change in momentum

$I = m (v_f-v_i)$

$I = 0.30\times (-4-3)$

$I = -2.1 kg.m/s$

Hence, impulse of the object is equal to $I = -2.1 kg.m/s$

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2 years ago

A basketball rolls to a stop along a gym floor without anyone touching it. What answer choice BEST explains why this occurred?

Minchanka [31]

Frictional forces are exerting force against the basketball, causing it to stop.

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2 years ago

Read 2 more answers

A charged particle is moving in a uniform magnetic field at a speed of 8.2Ã10^3 m/s in a direction 87Â° from the direction of th

77julia77 [94]

Answer:

Magnetic field, B = 0.042 T

Explanation:

It is given that,

Speed of charged particle, $v=8.2\times 10^3\ m/s$

Angle between velocity and the magnetic field, $\theta=87$

Charge, $q=5.7\ \mu C=5.7\times 10^{-6}\ C$

Magnetic force, F = 0.002 N

The magnetic force is given by :

$F=qvB\ sin\theta$

B is the magnetic field

$B=\dfrac{F}{qv\ sin\theta}$

$B=\dfrac{0.002}{5.7\times 10^{-6}\times 8.2\times 10^3\times sin(87)}$

B = 0.042 T

So, the strength of the magnetic field is 0.042 Tesla. Hence, this is the required solution.

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3 years ago

A 100 kg cart on a roller coaster has 1800J of kinetic energy. How fast is it going?<br> KE= 1/2mv2

34kurt

Answer:

the speed is equal to 6 m/s

Explanation:

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2 years ago

If the angular velocity of the pully is -8.4rad/s at a given time, and its anglar acceleration is -2.8rad/s2, what is the angula

snow_lady [41]

Here we know that

$\omega_i = - 8.4 rad/s$

$\alpha = - 2.8 rad/s^2$

$t = 1.5 s$

now from kinematics we have

$\omega_f = \omega_i + \alpha t$

now from above all values we have

$\omega_f = (-8.4 rad/s) + (-2.8 rad/s^2)(1.5)$

$\omega_f = -8.4 + (-4.2)$

$\omega_f = -12.6 rad/s$

so final angular speed is -12.6 rad/s

6 0

2 years ago