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Brrunno [24]
3 years ago
14

At what angle should you launch a projectile from the ground so that it has the

Physics
1 answer:
olga_2 [115]3 years ago
3 0
I'm guessing 45 degrees
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Which sentence provides evidence that heat travels from Earth’s mantle up to its crust?
olasank [31]

B. Earth’s outer surface is cooler than its interior layers.

Explanation:

  • The option given above is showing us that the temperature in the interior of the earth is higher than the temperature in the outer layer.
  • There is travel of heat from the inner core of the earth to the earth's crust. Due to the loss of heat when it reaches the outer layer, there arises a temperature difference.
  • The heat loss is due to the absorption of heat during its transfer. Hence, option B is the answer.
7 0
3 years ago
Two vectors are being added, one at an angle of 20.0 , and the other at 80.0. The only thing you know about the magnitudes is th
postnew [5]

Answer:a

Explanation: they are all positive

4 0
2 years ago
Virginia beach is 15 kilometers wide and 50 kilometers long. if 2 cm of rain falls on virginia beach, how many cubic meters of r
icang [17]
Assume that the shape of Virginia beach is rectangular.

Note that
1 km = 10³ m
1 cm = 10⁻² m

The area is
A = (15 km)*(50 km)
   = (15 x 10³ m)*(50 x 10³ m)
   = 7.5 x 10⁸ m²

Because 2 cm of rain fell, the volume is
V = (7.5 x 10⁸ m)*(2 x 10⁻² m) = 1.5 x 10⁶ m³

Answer:  1.5 x 10⁶ m³
4 0
3 years ago
During the experiment it is determined that, as the cart rolls between two points on the track, the work done on the cart by the
natita [175]

Answer:

Explanation:

If the work done on the cart is NET work

Then the work will result in an increase in kinetic energy

KE₀ + W = KE₁

½mv₀² + W = ½mv₁²

½(0.80)(0.61²) + 0.91 = ½(0.80)v₁²

v₁ = 1.626991...

v₁ = 1.6 m/s

4 0
2 years ago
50 grams of ice cubes at -15°C are used to chill a water at 30°C with mass mH20 = 200 g. Assume that the water is kept in a foam
Arada [10]

Answer : The final temperature is, 25.0^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of ice = 2.09J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of ice = 50 g

m_2 = mass of water = 200 g

T_f = final temperature = ?

T_1 = initial temperature of ice = -15^oC

T_2 = initial temperature of water = 30^oC

Now put all the given values in the above formula, we get:

50g\times 2.09J/g^oC\times (T_f-(-15))^oC=-200g\times 4.184J/g^oC\times (T_f-30)^oC

T_f=25.0^oC

Therefore, the final temperature is, 25.0^oC

5 0
3 years ago
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