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Karo-lina-s [1.5K]
3 years ago
9

Rupert (66 kg) is now in a full-pipe of radius 25 m, calculate the minimum speed at which Rupert can skate to ensure he will not

a fall off the skateboard at the top of the pipe?
**SHOW WORK**
Physics
1 answer:
V125BC [204]3 years ago
4 0

at the top most point if Rupert will not fall then normal force at the top point is almost zero for minimum speed

so here we can say

F_n + mg = m\frac{v^2}{R}

now if

F_n = 0

mg = \frac{mv^2}{R}

v = \sqrt{Rg}

v = \sqrt{25 \times 9.8}

v = 15.65 m/s

so above will be the minimum speed

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A particle of mass 10 g and charge 72 μC moves through a uniform magnetic field, in a region where the free-fall acceleration is
sineoko [7]

Answer:

-0.07163\hat k\ T or 0.07163 T into the page

Explanation:

m = Mass of particle = 10 g

a = Acceleration due to gravity = -9.8j m/s²

v = Velocity of particle = 19i km/s

q = Charge of particle = 72 μC

B = Magnetic field

Here the magnetic and gravitational forces on the particle are applied in the opposite direction so,

F_b=F_g

F_b=qvBsin\theta\\\Rightarrow F_b=qvBsin90\\\Rightarrow F_b=72\times 10^{-6}\times 19000B

F_g=ma\\\Rightarrow F_g=0.01\times -9.8

72\times 10^{-6}\times 19000B=0.01\times -9.8\\\Rightarrow B=\frac{0.01\times -9.8}{72\times 10^{-6}\times 19000}\\\Rightarrow B=-0.07163\hat k\ T

The magnetic field is 0.07163 T into the page

5 0
3 years ago
A 12000 kg train engine moving at 2.2 m/s hits and locks into 3 boxcars with a total mass of 25000 kg sitting still. If the coll
PilotLPTM [1.2K]

Answer:

V = 0.714m/s

Explanation:

Full solution calculation can be found in the attachment below.

From the principle of conservation of linear momentum, the sum of momentum before collision equals the sum of momentum after collision.

Before collision only the train had momentum. After the collision the train and the boxcars stick together and move as one body. The initial momentum of the train is now shared with the boxcars as they move together as one body. The both move with a common velocity v.

See the attachment below for the solution calculation.

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Suppose a person whose mass is m is being held up against the wall with a constant tangential velocity v greater than the minimu
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The specific heat of fat is roughly 1,700 J/(kgoC) whereas that for water is around 4200 J/(kgoC). A 0.63 kg solution of lipids
BartSMP [9]

Answer:

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

Explanation:

Total heat content of the fat = heat content of water +heat content of the lipids

Let it be Q

the Q= (mcΔT)_lipids + (mcΔT)_water

total mass of fat  M= 0.63 Kg

Q= heat supplied = 100 W in 5 minutes

ΔT= 20°C

c_lipid= 1700J/(kgoC)

c_water= 4200J/(kgoC)

then,

100\times5\times60= m(1700)20+(0.63-m)(4200)20

solving the above equation we get

m= 0.46 kg

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

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3 years ago
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