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kumpel [21]
4 years ago
9

Which three quantities can be used to calculate acceleration?

Physics
1 answer:
PtichkaEL [24]4 years ago
4 0
D is the correct answer, assuming that this is the special case of classical kinematics at constant acceleration. You can use the equation V = Vo + at, where Vo is the initial velocity, V is the final velocity, and t is the time elapsed. In D, all three of these values are given, so you simply solve for a, the acceleration.
A and C are clearly incorrect, as mass and force (in terms of projectile motion) have no effect on an object's motion. B is incorrect because it is not useful to know the position or distance traveled, unless it will help you find displacement. Even then, you would not have enough information to use a kinematics equation to find a.
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A proton moving at 3.90 106 m/s through a magnetic field of magnitude 1.80 T experiences a magnetic force of magnitude 7.20 10-1
VARVARA [1.3K]

Answer:

\theta=40^0

Explanation:

The magnitude of the magnetic force is

F_m=evB\sin\theta

To find the angle, we make \sin\theta subject of the formula

\implies \sin\theta=\frac{F_m}{evB}=\frac{7.20\times 10^{-13}}{1.6\times 10^{-19}\times 3.90\times 10^6\times 1.80}

\implies \sin\theta=0.641025641

\therefore \theta=\sin^{-1}=39.8683^0\\\implies \theta\approxeq 40^0

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How is the mass and speed of a particle related to its kinetic energy and temperature?
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Why will a set of stars appear at a different angle to us at different times of the year?
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A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target
stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

\theta=35^{\circ}

a.Let v_0 be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

v_x=v_0cos\theta=v_0cos35

v_y=v_0sin\theta=v_0sin35

x=v_0cos\theta\times t=v_0cos35\times t

t=\frac{30}{v_0cos35}

h=v_yt-\frac{1}{2}gt^2

Substitute the values

1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2

1.8=30tan35-\frac{6574.6}{v^2_0}

\frac{6574.6}{v^2_0}=21-1.8=19.2

v^2_0=\frac{6574.6}{19.2}

v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

t=\frac{30}{18.5cos35}

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0
3 years ago
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