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3 years ago

How should a line look like if an object is at rest(stopped) on a distance-time graph?

Physics

1 answer:

guajiro [1.7K]3 years ago

5 0

If an object is at rest, then its distance doesn't change as time goes on.

On the graph, the points for each time are all at the same distance. They form a horizontal line.

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An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary

Talja [164]

Answer:

A) 5.2 x 10³ N

B) 8.8 x 10³ N

Explanation:

Part A)

$F_{g}$ = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

$T$ = Tension force in upward direction

$F_{d}$ = Drag force in upward direction = 1800 N

Force equation for the motion of craft is given as

$F_{g}$ - $F_{d}$ - $T$ = 0

7000 - 1800 - $T$ = 0

$T$ = 5200 N

$T$ = 5.2 x 10³ N

Part B)

$F_{g}$ = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

$T$ = Tension force in upward direction

$F_{d}$ = Drag force in downward direction = 1800 N

Force equation for the motion of craft is given as

$T$ - $F_{g}$ - $F_{d}$ = 0

$T$ - 7000 - 1800 = 0

$T$ = 8800 N

$T$ = 8.8 x 10³ N

4 0

4 years ago

Indicate whether the statement is true or false. an astronaut weighs the same on earth as in space.

tino4ka555 [31]

Weight is different (but mass is the same)

6 0

3 years ago

Read 2 more answers

In a 350-m race, runner A starts from rest and accelerates at 1.6 m/s^2 for the first 30 m and then runs at constant speed. Runn

kifflom [539]

Answer:

B can take 0.64 sec for the longest nap .

Explanation:

Given that,

Total distance = 350 m

Acceleration of A = 1.6 m/s²

Distance = 30 m

Acceleration of B = 2.0 m/s²

We need to calculate the time for A

Using equation of motion

$s=ut+\dfrac{1}{2}at_{A}^2$

Put the value in the equation

$30=0+\dfrac{1}{2}\times1.6\times t_{A}^2$

$t_{A}=\sqrt{\dfrac{30\times2}{1.6}}$

$t_{A}=6.12\ sec$

We need to calculate the time for B

Using equation of motion

$s=ut+\dfrac{1}{2}at_{B}^2$

Put the value in the equation

$30=0+\dfrac{1}{2}\times2.0\times t_{B}^2$

$t_{B}=\sqrt{\dfrac{30\times2}{2.0}}$

$t_{B}=5.48\ sec$

We need to calculate the time for longest nap

Using formula for difference of time

$t'=t_{A}-t_{B}$

$t'=6.12-5.48$

$t'=0.64\ s$

Hence, B can take 0.64 sec for the longest nap .

4 0

4 years ago

What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?

Nastasia [14]

Answer:

$F=1.26*10^{-3}N$

Explanation:

Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

$F=\frac{kq_1q_2}{d^2}$

We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges( $q_1,q_2$ ) and inversely proportional to the square of the distance(d) that separates them.

Replacing the given values, where k is the Coulomb constant:

$F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-30*10^{-9}C)(-30*10^{-9}C)}{(8*10^{-2}m)^2}\\F=1.26*10^{-3}N$

8 0

3 years ago

Which of the following materials could have been found in the giant cloud that formed the solar system?

iren2701 [21]

Liquid and solid water were not in the giant gas cloudr

6 0

3 years ago

Read 2 more answers