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solong [7]
3 years ago
11

What will happen to a spacecraft that is moving far out in space, and that is outside Earth’s atmosphere and the pull of gravity

?
A. It will continue to move at the same speed and in the same direction.
B. It will eventually slow down and fall to Earth.
C. It will continue to move faster and faster away from Earth.
D. It will burn up due to friction.
Physics
1 answer:
Akimi4 [234]3 years ago
4 0
I think the answer is D
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An irregular object of mass 3 kg rotates about an axis, about which it has a radius of gyration of 0.2 m, with an angular accele
Citrus2011 [14]

Answer:

The magnitude of the applied torque is 6.0\times10^{-2}\ N-m

(e) is correct option.

Explanation:

Given that,

Mass of object = 3 kg

Radius of gyration = 0.2 m

Angular acceleration = 0.5 rad/s²

We need to calculate the applied torque

Using formula of torque

\tau=I\times\alpha

Here, I = mk²

\tau=mk^2\times\alpha

Put the value into the formula

\tau=3\times(0.2)^2\times0.5

\tau=0.06\ N-m

\tau=6.0\times10^{-2}\ N-m

Hence, The magnitude of the applied torque is 6.0\times10^{-2}\ N-m

7 0
4 years ago
A small but measurable current of 5.8 × 10-10 A exists in a copper wire whose diameter is 3.0 mm. The number of charge carriers
swat32

Answer:

a) The current density ,J = 2.05×10^-5

b) The drift velocity Vd= 1.51×10^-15

Explanation:

The equation for the current density and drift velocity is given by:

J = i/A = (ne)×Vd

Where i= current

A = Are

Vd = drift velocity

e = charge ,q= 1.602 ×10^-19C

n = volume

Given: i = 5.8×10^-10A

Raduis,r = 3mm= 3.0×10^-3m

n = 8.49×10^28m^3

a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]

J = (5.8×10^-10) /(2.83×10^-5)

J = 2.05 ×10^-5

b) Drift velocity, Vd = J/ (ne)

Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)

Vd = (2.05×10^-5)/(1.36 ×10^10)

Vd = 1.51× 10^-5

8 0
3 years ago
Read 2 more answers
As an object sinks in a fluid, the buoyant force ____.
kiruha [24]
Beginning when the bottom of the object first touches the water,
and as it descends and more and more of it goes under, the
buoyant force on it increases during that time.

As soon as the object is completely underwater, it doesn't matter
how deep under it is, the buoyant force on it remains the same.

3 0
3 years ago
Read 2 more answers
Alan leaves Los Angeles at 8:00 a.m. to drive to San Francisco, 400 mi away. He travels at a steady 45.0 mph . Beth leaves Los A
WARRIOR [948]

Answer:

a.Beth

b.2232 s

Explanation:

We are given that

Distance,d=400 mi

Speed of Alan,v=45 mph

Speed of Beth,v'=55 mph

a.Time =\frac{distance}{speed}

Using the formula

Time taken by Alan=\frac{400}{45}=8.89 hr

Time taken by Beth=\frac{400}{55}=7.27hr

Alan will reach San Francisco at 4:53 PM

Beth will reach San Francisco at 4:16 PM

Beth will reach before Alan.

b.Difference between time=8.89-7.27=1.62 hr

t=1.62 hr

1.62-1=0.62 hr

0.62 hr=0.62\times 60\times 60=2232 s

Hence, Beth has to wait 2232 s for Alan to arrive .

6 0
3 years ago
Is the following chemical equation balanced?<br><br> MgI2 + Br2 MgBr2 + I2 <br><br> yes<br><br> no
Brums [2.3K]

Yes

Explanation:

This chemical equation:

Mg I_2 + Br_2 \rightarrow Mg Br_2 + I_2

is balanced, because the number of atoms of each element is the same on the reactant side and on the product side. In fact:

- Mg: one atom on the left, one on the right

- I: 2 atoms on the left, and 2 on the right

- Br: 2 atoms on the left, and 2 on the right

So, the reaction is balanced.

3 0
3 years ago
Read 2 more answers
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