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Murrr4er [49]
2 years ago
13

What happens when a temperature increases.​

Physics
2 answers:
Liono4ka [1.6K]2 years ago
5 0
The density decreases and convection causes the hot air particles to rise! BRANLIEST
Maksim231197 [3]2 years ago
5 0

Answer:

if the tempertaure is increase it effects us very badly. it affects not only us . it effects our whole enviroment.

You might be interested in
A motion sensor emits sound, and detects an echo 0.0115 s after. A short time later, it again emits a sound, and hears an echo a
Mekhanik [1.2K]

Answer:

1.17 m

Explanation:

From the question,

s₁ = vt₁/2................ Equation 1

Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.

Given: v = 343 m/s, t = 0.0115 s

Substitute into equation 1

s₁ = (343×0.0115)/2

s₁ = 1.97 m.

Similarly,

s₂ = vt₂/2.................. Equation 2

Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo

Given: v = 343 m/s, t₂ = 0.0183 s

Substitute into equation 2

s₂ = (343×0.0183)/2

s₂ = 3.14 m

The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁

s₂-s₁ =  (3.14-1.97) m = 1.17 m

7 0
3 years ago
Read 2 more answers
Suppose a baseball pitcher throws the ball to his catcher.
amm1812

a) Same

b) Same

c) Same

d) Throw the ball takes longer

e) F is larger when the ball is catched

Explanation:

a)

The change in speed of an object is given by:

\Delta v = |v-u|

where

u is the initial velocity of the object

v is the final velocity of the object

The change in speed is basically the magnitude of the change in velocity (because velocity is a vector, while speed is a scalar, so it has no direction).

In this problem:

- In situation 1 (pitcher throwing the ball), the initial velocity is

u = 0 (because the ball starts from rest)

while the final velocity is v, so the change in speed is

\Delta v=|v-0|=|v|

- In situation 2 (catcher receiving the ball), the initial velocity is now

u = v

while the final velocity is now zero (ball coming to rest), so the change in speed is

\Delta v =|0-v|=|-v|

Which means that the two situations have same change in speed.

b)

The change in momentum of an object is given by

\Delta p = m \Delta v

where

m is the mass of the object

\Delta v is the change in velocity

If we want to compare only the magnitude of the change in momentum of the object, then it is given by

|\Delta p|=m|\Delta v|

- In situation 1 (pitcher throwing the ball), the change in momentum is

\Delta p = m|\Delta v|=m|v|=mv

- In situation 2 (catcher receiving the ball), the change in momentum is

\Delta p = m\Delta v = m|-v|=mv

So, the magnitude of the change in momentum is the same (but the direction is opposite)

c)

The impulse exerted on an object is equal to the change in momentum of the object:

I=\Delta p

where

I is the impulse

\Delta p is the change in momentum

As we saw in part b), the change in momentum of the ball in the two situations is the same, therefore the impulse exerted on the ball will also be the same, in magnitude.

However, the direction will be opposite, as the change in momentum has opposite direction in the two situations.

d)

To compare the time of impact in the two situations, we have to look closer into them.

- When the ball is thrown, the hand "moves together" with the ball, from back to ahead in order to give it the necessary push. We can verify therefore that the time is longer in this case.

- When the ball is cacthed, the hand remains more or less "at rest", it  doesn't move much, so the collision lasts much less than the previous situation.

Therefore, we can say that the time of impact is longer when the ball is thrown, compared to when it is catched.

e)

The impulse exerted on an object can also be rewritten as the product between the force applied on the object and the time of impact:

I=F\Delta t

where

I is the impulse

F is the force applied

\Delta t is the time of impact

This can be rewritten as

F=\frac{I}{\Delta t}

In this problem, in the two situations,

- I (the impulse) is the same in both situations

- \Delta t when the ball is thrown is larger than when it is catched

Therefore, since F is inversely proportional to \Delta t, this means that the force is larger when the ball is catched.

6 0
3 years ago
If the kinetic energy of an electron is 4.1e-18 j, what is the speed of the electron? (you can use the approximate (nonrelativis
arlik [135]
The kinetic energy of the electron is
K= \frac{1}{2}mv^2
where m=9.1 \cdot 10^{-31} kg is the mass of the electron and v its speed. Since we know the value of the kinetic energy, K=4.1 \cdot 10^{-18} J, we can find the value of the speed v:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2\cdot 4.1 \cdot 10^{-18}J}{9.1 \cdot 10^{-31}kg} }  = 3\cdot 10^6 m/s
3 0
3 years ago
three carts of masses 2 kg, 18 kg, and 9 kg move on a frictionless horizontal track with speeds of 10m/s, 8m/s, and 2m/s. the ca
Elan Coil [88]

Answer:

V = 6.3 m/s

Explanation:

Given:

m₁ = 2 kg

m₂ = 18 kg

m₃ = 9 kg

V₁ = 10 m/s

V₂ = 8 m/s

V₃ = 2 m/s

__________

V - ?

Let us write the momentum conservation law for an inelastic impact:

m₁·V₁ + m₂·V₂ + m₃·V₃ = (m₁ +m₂ + m₃) ·V

Cart speed after interaction:

V = ( m₁·V₁ + m₂·V₂ + m₃·V₃ ) / (m₁ +m₂ + m₃)

V = (2·10 + 18·8 + 9·2) / ( 2 + 18 + 9) = 182 / 29 ≈  6.3 m/s

4 0
1 year ago
Can someone help me ?
hammer [34]

Answer:

1)    Time interval                 Blue Car                      Red Car

          0 - 2 s                Constant Velocity           Increasing Velocity

          2 - 3 s                Constant Velocity           Constant Velocity

          3 - 5 s                Constant Velocity           Increasing Velocity

          5 - 6 s                Constant Velocity           Decreasing Velocity

2) For Red and Blue car y₂  = 120       v = \frac{y_{2}-y_{1}}{t_{2}-t_{1}} = \frac{120-0}{6-0} = 20 m/s

     We get the same velocity for two cars because it is the average velocity of the car at the given interval of time. It is measured for initial and final position.

3)   At t = 2s, the cars are the same position, and are moving at the same rate

                    Position - same

                    Velocity - same

The position-time graph shares the same spot for two cars.

4 0
3 years ago
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