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slamgirl [31]
3 years ago
12

What would the answer be ?

Physics
1 answer:
Tpy6a [65]3 years ago
4 0

Answer:

im going between 2 of them b and c but i would have choose b

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A rocket in orbit just above the atmosphere is moving in uniform circular motion. The radius of the circle in which it moves is
Zinaida [17]
Centripetal acceleration is given by the formula

a_{cen}= \frac{v^2}{r}

Since we know the value of the acceleration is 9.8 m/s² and we know r, we can solve for the speed, which is the magnitude of v.  Note the value of a is just the acceleration due to gravity, which is what makes the rocket orbit and not fly off into space.

9.8 \frac{m}{s^2}=\frac{v^2}{6.381 \times 10^6m}   \\  \\ 9.8 \frac{m}{s^2} \times 6.381 \times 10^6m=v^2 \\  \\ v= \sqrt{9.8 \frac{m}{s^2} \times 6.381 \times 10^6m} =24755 \frac{m}{s}
8 0
3 years ago
A 33.0 g iron rod, initially at 22.7 ∘c, is submerged into an unknown mass of water at 63.3 ∘c, in an insulated container. the f
professor190 [17]
When the iron and the water reach thermal equilibrium, they have same temperature, T=58.5^{\circ}C.
We can consider this as an isolated system, so the heat released by the water is equal to the heat absorbed by the iron.

The hear released by the water is:
Q_w =m_w C_{sw} \Delta T_w
where m_w is the water mass, C_{sw}=4.186 J/(g^{\circ}C) is the specific heat of the water, and \Delta T_w = 63.6^{\circ}-58.5^{\circ}=5.1^{\circ}C is the variation of temperature of the water.

Similarly, the heat absorbed by the iron is:
Q_i = m_i C_{si} \Delta T_i
where m_i = 33.0 g is the iron mass, C_{si}=0.444 J/(g^{\circ}C) is the iron specific heat, and \Delta T_i = 58.5^{\circ}-22.7^{\circ}=35.8^{\circ}C is the variation of temperature of the iron.

Writing Q_w=Q_i and replacing the numbers, we can solve to find mw, the mass of the water:
m_w= \frac{m_i C_{si} \Delta T_i}{C_{sw} \Delta T_w} =24.6 g
4 0
3 years ago
From the ground an object is vertically thrown upwards with an angle of theta.
Mashutka [201]

Answer:

u/2 √(1 + 3 cos² θ)

Explanation:

The object is thrown at an angle θ, so the velocity has two components, vertical and horizontal.

Initially, the vertical component is u sin θ and the horizontal component is u cos θ.

At the maximum height, the vertical component is 0 and the horizontal component is u cos θ.

The mean vertical velocity is:

(u sin θ + 0) / 2 = u/2 sin θ

The mean horizontal velocity is:

(u cos θ + u cos θ) / 2 = u cos θ

The net mean velocity can be found with Pythagorean theorem:

v² = (u/2 sin θ)² + (u cos θ)²

v² = u²/4 sin² θ + u² cos² θ

v² = u²/4 (1 − cos² θ) + u² cos² θ

v² = u²/4 (1 − cos² θ) + u²/4 (4 cos² θ)

v² = u²/4 (1 − cos² θ + 4 cos² θ)

v² = u²/4 (1 + 3 cos² θ)

v = u/2 √(1 + 3 cos² θ)

8 0
4 years ago
What do repeated trails in an experiment allow scientists to do?
fredd [130]

Answer:

Explanation:

When we do multiple trials of the same experiment, we can make sure that our results are consistent and not altered by random events. Multiple trials can be done at one time. If we were testing a new fertilizer, we could test it on lots of individual plants at the same time.

3 0
3 years ago
a moving electron accelerates at 5200m/s^2 in a 55.0 direction. After 0.350as,it has a velocity of 6598m/s in a -20.5 direction.
Lady bird [3.3K]

Answer:

  -3802 m/s

Explanation:

The y-component of the final velocity is ...

  (6598 m/s)·sin(-20.5°) ≈ -2310.7 m/s

The y-component of the velocity due to acceleration is ...

  (5200 m/s²)(0.350 s)sin(55°) ≈ 1490.9 m/s

Then the initial velocity in the y-direction is found from ...

  initial velocity + change in velocity = final velocity

  initial velocity = (final velocity) - (change in velocity)

  = -2310.7 m/s - 1490.9 m/s ≈ -3802 m/s

6 0
4 years ago
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