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Shtirlitz [24]
2 years ago
7

An unmanned spacecraft leaves for venus. Which statements about the spacecraft's journey are true?

Physics
2 answers:
olganol [36]2 years ago
8 0
D. The mass remains the same
zvonat [6]2 years ago
3 0
These are the correct answers<span>

A. The weight of the spacecraft keeps changing. 
~ Weight is dependent on gravity which IS changing.
</span>
<span>D. The mass of the spacecraft remains the same. 
~ Mass doesn't change

*Answer for Plato Users </span>
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The diagram shows a heater above a thermometer. The thermometer bulb is in the position shown. Which row shows how the heat ener
balu736 [363]

Answer:

The diagram shows a heater above a thermometer. The thermometer bulb is in the position shown. How the heat

5 0
2 years ago
Adam observed properties of four different waves and recorded observations about the frequency and volume of each one in his cha
butalik [34]
I believe this answer is B but my believing sucks so
3 0
2 years ago
A particle decelerates uniformly from a speed of 30 cm/s to rest in a time interval of 5.0 s. It then has a uniform acceleration
wel

Answer:

V=20cm/s

Explanation:

The average speed is the distance total divided the time total:

V=X/T

First stage:

T1=5s

v_{f}  =v_{o} - at

But, v_{f}  =0   (decelerates to rest)

then: a =v_{o} /t=0.3/5=0.06m/s^{2}

on the other hand:

x =v_{o}*t - 1/2*at^{2}=0.3*5-1/2*0.06*5^{2}=0.75m

X1=75cm

Second stage:

T2=5s

x =v_{o}*t + 1/2*at^{2}=0+1/2*0.1*5^{2}=1.25m

X2=125cm

Finally:

X=X1+X2=200cm

T=T1+T2=10s

V=X/T=20cm/s

8 0
2 years ago
In a certain clock, a pendulum of length L1 has a period T1 = 0.95s. The length of the pendulum
gulaghasi [49]

Answer:

Ratio of length will be \frac{L_2}{L_1}=1.108

Explanation:

We have given time period of the pendulum when length is L_1 is T_1=0.95sec

And when length is L_2 time period T_2=1sec

We know that time period is given by

T=2\pi \sqrt{\frac{L}{g}}

So 0.95=2\pi \sqrt{\frac{L_1}{g}}----eqn 1

And 1=2\pi \sqrt{\frac{L_2}{g}}-------eqn 2

Dividing eqn 2 by eqn 1

\frac{1}{0.95}=\sqrt{\frac{L_2}{L_1}}

Squaring both side

\frac{L_2}{L_1}=1.108

8 0
2 years ago
A projectile is launched at an angle of 60° from the horizontal and at a velocity of
gayaneshka [121]

Answer:

60*12.0= 720 = v/60 * 12.0 squared which is 1,728

Explanation:

Horizontal velocity component: Vx = V * cos(α)

5 0
2 years ago
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