Select the decimal that is equivalent to 29/37

A 70.0Kg cyclist Rides a 15.0kg

Inessa [10]

Answer:

$1275 kg*m/s$

Explanation:

We'll use the momentum equation:

$p=mv$

where:

p = momentum

m = mass

v = velocity

Since we're doing the magnitude of momentum of the system, we'll add the mass of the cyclist and the mountain bike together:

$70.0 kg + 15.0kg=85.0 kg$

Given that, we can now substitute our given values into the momentum equation:

$p=mv\\p=85.0kg*15.0m/s$

Our final answer is:

$p= 1275 kg*m/s$

4 0

2 years ago

Uranium-238 eventually decays into

GalinKa [24]

Uranium-238 decays by alpha emission into thorium-234, which itself decays by beta emission to protactinium-234, which decays by beta emission to uranium-234, and so on. The various decay products, (sometimes referred to as “progeny” or “daughters”) form a series starting at uranium-238.

4 0

3 years ago

If the tire reaches a temperature of 48 ∘c, what fraction of the original air must be removed if the original pressure of 250 kp

andrey2020 [161]

Assuming air as ideal gas and amount of air in no of moles is known then by gas law,

PV= nRT

Pressure is constant

P* (change in volume) = nR* (change in temperature)

7 0

3 years ago

Indigenous people sometimes cooked in watertight baskets by placing enough hot rocks into the water to bring it to a boil. What

yaroslaw [1]

Answer:

The rock has a mass of 4.02 kg

Explanation:

Step 1: Data given

Mass of the rock = TO BE DETERMINED

Temperature of the rock = 500 °C

Mass of the water =4.24 kg

⇒ loses 0.044kg as vapor

Initial temperature of the water = 29°C

Final temperature = 100°C

Specific heat of rock = 0.20 kcal/kg °C

Specific heat of water = 1kcal/kg°C

Latent heat of vaporization = 539 kcal/kg

Step 2: formules

Qlost,rock + Qgained,water = 0

Qtotal,water = Qwater +Qvapor

Step 3: Calculate Qvapor

Qvapor = mass of vapor * Latent heat of vapor

Qvapor = 0.044kg * 539 kcal/kg = 23.716 kcal

Step 4: Calculate Qwater

Qwater = mass of water * specific heat * Δtemperature

Qwater = 4.196 kg * 1kcal/kg°C *( 100-29)

Qwater = 297.916 kcal

Step 5: Calculate Qwater,total

Qwater,total = Qwater + Qvapor

Qwater,total = 23.716 kcal + 297.916 = 321.632 kcal

Step 6: Calculate Qrock

Qrock = mass of rock * specific heat rock * Δtemperature

Qrock = mass of rock * 0.20 kcal/kg°C * (100-500)

Qrock = mass of rock * -80 kcal/kg

Step 7: Calculate mass of rock

Qlost,rock + Qgained,water = 0

Qlost,rock = -Qgained,water

mass of rock * -80 kcal/kg = -321.632 kcal

mass of rock = 4.02 kg

The rock has a mass of 4.02 kg

7 0

2 years ago

The specific heat of water is 4,186 J/kg.'C. Approximately how much heat must

Artyom0805 [142]

Explanation:

Q= mc∆T

∆T= 5-24=- 19

Q= 0.5*4186*-19

Q= -39767 J

negative sign show heat releases

4 0

2 years ago

Select the decimal that is equivalent to 29/37​

Select the decimal that is equivalent to 29/37