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3 years ago

Please answer the question below

Physics

1 answer:

scoundrel [369]3 years ago

6 0

The correct answer is B.
In a velocity vs time graph, a line going up means an increase in velocity, or speed, which correlates to positive acceleration. Negative acceleration means slowing down.

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What is a “plate” in geological terms

Tju [1.3M]

Answer:a plate is a rigid segment Of The earth's lithosphere that moves horizontally and adjoins other plates along then zone of seismic activity.

Explanation:

A plate is a rigid segment Of The earth's lithosphere that moves horizontally and adjoins other plates along then zone of seismic activity.plates may include portion of both continents and ocean basin.

3 0

3 years ago

Force due to gravity. While JWST is in orbit, the Earth will be at a distance of 1.494 x 109 m from the telescope, and the Sun w

Alona [7]

Answer:

the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

Explanation:

Given that;

distance of earth from the telescope r1 = 1.494 × 10⁹ m

and the Sun will be 1.49598 × 10¹¹ m further away

so r2 = r1 + 1.49598 × 10¹¹

r2 = 1.494 × 10⁹ m + 1.49598 × 10¹¹ m = 1.51092 × 10¹¹ m

mass of sun Ms = 1.9884 × 10³⁰ kg

mass of earth Me = 5.945× 10²⁴ kg

mass of JWST Mj = 6500 kg

What is the gravitational force JWST will feel from the Sun (strength and direction)?

the gravitational force of the sun will be attractive based on Newton law of gravitational force; so

Fjs = GMjMs / r2²

constant G = 6.674 × 10⁻¹¹ Nm²/kg²

Force on the JWST by the sun will be;

Fjs = GMjMs / r2² { leftward}

we substitute

Fjs = [(6.674 × 10⁻¹¹ Nm²/kg²)(6500 kg )(1.9884 × 10³⁰ kg)] / (1.51092 × 10¹¹ m)²

= (8.62587804 × 10²³) / ( 2.28287925 × 10²² )

= 37.785 Newton { leftward }

Therefore, the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

7 0

3 years ago

A ship sets sail from Rotterdam, The Netherlands, heading due north at 8.00 m/s relative to the water. The local ocean current i

Nuetrik [128]

Answer:

The velocity of the ship relative to the earth V = 9.05 $\frac{m}{s}$

Explanation:

The local ocean current is = 1.52 m/s

Direction $\theta$ = 40°

Velocity component in X - direction $V_{x}$ = 1.52 $\cos 40$ °

$V_{x}$ = 1.164 $\frac{m}{s}$

Velocity component in Y - direction $V_{y}$ = 8 + 1.52 $\sin 40$ °

$V_{y}$ = 8.97 $\frac{m}{s}$

The velocity of the ship relative to the earth

$V = \sqrt{V_{x}^{2} + V_{y} ^{2} }$

Put the values of $V_{x}$ and $V_{y}$ we get,

⇒ $V = \sqrt{1.164^{2} + 8.97 ^{2} }$

⇒ V = 9.05 $\frac{m}{s}$

This is the velocity of the ship relative to the earth.

7 0

3 years ago

In comparison with a steep slope, a gentle slope _____.

kifflom [539]

A gentle slope requires less force over a longer distance as compared to steep slope.

Explanation:

Mechanical advantage of a slope is equal to the ratio of length of slope and the height. A steep slope has shorter length as compared to a gentle slope for the same height. Therefore, mechanical advantage of a gentle slope is more than that of a steep slope. Hence, a gentle slope requires less force over a long distance than a steep slope.

6 0

3 years ago

Read 2 more answers

Calculate the pressure exerted on the floor by the boy standing on both feet if the weight of the boy is 40kg. Assume that the a

Jlenok [28]

Answer:

P = 1333.33 N

Explanation:

The pressure exerted by the boy on the floor can be calculated by the following equation:

$P = \frac{F}{A}$