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3 years ago

Select all the correct ANSWERS.

Physics

1 answer:

MissTica3 years ago

3 0

The ground exerts an equal force on the golf ball.

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PLEASE ANSWER WILL GIVE BRAINLIEST

Karolina [17]

Answer:

B. NET force: 2 resultant motion: left

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C. Net force: 3 Resultant motion: Left

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D. Net Force: 7 Resultant motion: right

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E. Net Force:0 resultant motion: NO MOTION

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F. NET Force: 3 resultant motion: Down

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G. NET FORCE: 10 resultant motion: up

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H. Net force: 3 Resultant motion: left

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I. Net force: 50 Resultant motion: right

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J. NET FORCE: 75 Resultant motion: down

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K. Net force :200 Resultant motion: Right

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L. Net force: 0 resultant motion:No motion

Explanation:

8 0

2 years ago

A woman launches a boat from one shore of a straight river and wants to land at the point directly on the opposite shore. If the

I am Lyosha [343]

Answer:

If she stands on the North side of a river flowing to the East at 5 mph,

she must head towards the SouthWest to arrive on the South side of the river directly across from her starting point and we have

x^2 + 5^2 = 10^2 where x is her speed directly across the river

x = (75)^1/2 = 8.66 mph towards the South

sin theta = 5 / 10 = 1/2

She must angle the boat at 30 deg from straight South

4 0

1 year ago

What kind of medium does ocean waves have

Anika [276]

Ocean waves propagate through the medium called 'sea water'.

3 0

2 years ago

Particle q₁ has a charge of 2.7 μC and a velocity of 773 m/s. If it experiences a magnetic force of 5.75 × 10⁻³ N, what is the s

Ne4ueva [31]

The intensity of the magnetic force F experienced by a charge q moving with speed v in a magnetic field of intensity B is equal to
$F=qvB \sin \theta$
where $\theta$ is the angle between the directions of v and B.

1) Re-arranging the previous formula, we can calculate the value of the magnetic field intensity. The charge is $q=2.7 \mu C=2.7 \cdot 10^{-6}C$ . In this case, v and B are perpendicular, so $\theta=90^{\circ}$ , therefore we have:
$B= \frac{F}{qv \sin \theta} = \frac{5.75 \cdot 10^{-3}N}{(2.7 \cdot 10^{-6}C)(773m/s)\sin 90^{\circ}}=2.8 T$

2) In this second case, the angle between v and B is $\theta=55^{\circ}$ . The charge is now $q=42.0 \mu C=42.0 \cdot 10^{-6}C$ , and the magnetic field is the one we found in the previous part, B=2.8 T, so we can find the intensity of the force experienced by this second charge:
$F=qvB \sin \theta=(42\cdot 10^{-6}C)(1.21 \cdot 10^3 m/s)(2.8 T)(\sin 55^{\circ})=0.12 N$

5 0

2 years ago

Read 2 more answers

Light passing through the center of a lens will carry on undeviated.

icang [17]

I think the answer is B true

4 0

2 years ago