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Katen [24]
3 years ago
9

In a position time graph, a curved line represents: acceleration no movement constant velocity

Physics
1 answer:
xz_007 [3.2K]3 years ago
8 0

A curved line on a position/time graph shows that the speed is changing.

So right there, we know there is acceleration.

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True or false? To develop good alternatives, one should brainstorm ideas and consider different perspectives
Contact [7]
From my experience, I would say it is true.
7 0
3 years ago
A 0.5 kg ball moves in a circle that is 0.4 m in radius at a speed of 4.0 m/s. Calculate the force exerted on the ball.
blsea [12.9K]

Answer:

Explanation:

Given a ball of mass m= 0.5kg

The ball moves in as circle of radius r= 0.4m

Speed of the ball is v = 4m/s

Centripetal force is exerted on ball and it is given as

Fc = m•ac

ac is centripetal acceleration and it is given as

ac = v²/r

Then,

Fc = mv²/r

Fc = 0.5 × 4²/0.4

Fc = 20N.

The force exerted on the ball is 20N

5 0
3 years ago
What is the effect of pressure on the solubility of gases in liquids?
34kurt
The solubility of gases in liquids increases with the increase in pressure.
3 0
3 years ago
A bolt of lightning discharges 9.7 C in 8.9 x 10^-5 s. What is the average current during the discharge?
Anastaziya [24]

Answer: 1.089\times 10^5\ A

Explanation:

Given

Charge discharged Q=9.7\C

time taken t=8.9\times 10^{-5}\ s

Current is given as rate of change of discharge i.e.

\Rightarrow I=\dfrac{Q}{t}\\\\\Rightarrow I=\dfrac{9.7}{8.9\times 10^{-5}}\\\\\Rightarrow I=1.089\times 10^5\ A

Therefore, the average current is 1.089\times 10^5\ A

3 0
2 years ago
An insulated beaker with negligible mass contains 0.250 kg of water at 75.0C. How many kilograms of ice at -20.0C must be droppe
kkurt [141]

Answer:

The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg

Explanation:

Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water

Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C

To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.

Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C

Latent heat of ice = L = 334000 J/kg

Specific heat capacity of water = C = 4186 J/kg.°C

Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m

Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J

543600 m = 36627.5

m = 0.0674 kg = 67.4 g of ice.

3 0
3 years ago
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