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Verizon [17]
3 years ago
7

Nitrogen condenses at -196 °C and freezes at -210 °C. At which of the

Physics
1 answer:
maksim [4K]3 years ago
5 0

Answer:

-200°C

Explanation:

This is because at -200°C the temperature is below -196°C and above -210°C .Thus Nitrogen can be in liquid state.

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A crowbar having the length of 1.75m is used to balance a load between the fulcrum and the load is 0.5m calculate MA
aliina [53]

Answer:

1.70

Explanation:

substract 1.75-0.5

3 0
2 years ago
A rocket starting from its launch pad is subjected to a uniform acceleration of 100 meters/second2. Determine the time needed to
gizmo_the_mogwai [7]

Answer:

10s

Explanation:

Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.

If acceleration is constant, then the velocity is changing by a constant amount.

With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:

v=at+v_o where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.

v=at+v_o

(1000\text{ [m/s]})=(100 \text{ } [\text{m/s}^2] )t+(0\text{ [m/s]})

1000\text{ [m/s]}=100 \text{ } [\text{m/s}^2] *t

\dfrac{1000\text{ [m/s]}}{100 \text{ } [\text{m/s}^2]}=\dfrac{100 \text{ } [\text{m/s}^2] *t}{100 \text{ } [\text{m/s}^2]}

10\text{ [s]}=t

So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.

<u>Verification:</u>

In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.

Recognizing that the acceleration is a=\dfrac{100 [\frac{m}{s}]}{1[s]}, the velocity increases by 100 units [m/s] every second.

At time 0[s], the velocity is 0[m/s]

At time 1[s], the velocity is 100[m/s]

At time 2[s], the velocity is 200[m/s]

At time 3[s], the velocity is 300[m/s]

At time 4[s], the velocity is 400[m/s]

At time 5[s], the velocity is 500[m/s]

At time 6[s], the velocity is 600[m/s]

At time 7[s], the velocity is 700[m/s]

At time 8[s], the velocity is 800[m/s]

At time 9[s], the velocity is 900[m/s]

At time 10[s], the velocity is 1000[m/s]

So, indeed, after 10 seconds, the velocity reaches 1000 m/s

5 0
2 years ago
What happens when fusion begins in the outer shell of a star? The energy produced causes the star to contract into a dwarf. The
Aleks [24]

Answer:

D

Explanation:

4 0
2 years ago
A meter stick moves parallel to its axis with speed of 0.96 c relative to you. What would you measure for the length of the stic
Fed [463]

Answer:

The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

Explanation:

Given that,

Relative speed of stick v= 0.96 c

Speed of light c= 2.99793\times10^{8}\ m/s

Proper length of stick = 1 m

We need to calculate the length of the stick

Using formula of length

\Delta l=\Delta l_{0}\sqrt{(1-\dfrac{v^2}{c^2})}

Put the value into the formula

\Delta l=1\sqrt{1-\dfrac{(0.96)^2c^2}{c^2}}

\Delta l=1\sqrt{1-(0.96)^2}

\Delta l=0.28\ m

We need to calculate the time the stick take to move

Using formula of time

t=\dfrac{\Delta l}{v}

Put the value into the formula

t=\dfrac{0.28}{0.96\times(2.99793\times10^{8})}

t=9.72\times10^{-10}\ sec

t=0.97\ ns

Hence, The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

7 0
2 years ago
A potentialiterence of 12 volts is induced across a straight wire 0.20 meters long as it is moved at a constant speed of 3.0 met
shtirl [24]

Strength of the magnetic field: 20 T

Explanation:

For a conductive wire moving perpendicular to a magnetic field, the electromotive force (voltage) induced in the wire due to electromagnetic induction is given by

\epsilon=BvL

where

B is the strength of the magnetic field

v is the speed of the wire

L is the length of the wire

For the wire in this problem, we have:

\epsilon=12 V (induced emf)

L = 0.20 m (length of the wire)

v = 3.0 m/s (speed)

Solving  for B, we find the strength of the magnetic field:

B=\frac{\epsilon}{vL}=\frac{12}{(0.20)(3.0)}=20 T

Learn more about magnetic fields:

brainly.com/question/3874443

brainly.com/question/4240735

#LearnwithBrainly

6 0
3 years ago
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