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3 years ago

13

A bicyclist with a mass of 50 kg is traveling at a rate of 30 m/s. It accelerates to a rate of 50 m/s in 5 seconds. What is the

net force acting on the bicycle? *

1 answer:

balandron [24]3 years ago

8 0

Answer:

F=m*(v^2/r)

F=82*(8^2/30)

F=174.9N

Explanation:

brainlest pls

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To increase the gravitational force between the two objects above, I could

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Option A

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2 years ago

a bus travels 4 km due north and 3 km due west going from bus station a to bus station b. the magnitude of the bus displacement

ladessa [460]

Answer:

5km

Explanation:

Magnitude of displacement is found by getting the resultant. Resultant is same as the bypotenuse hence

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3 years ago

A force of 3600 N is exerted on a piston that has an area of 0.030 m2. What force is exerted on a second piston that has an area

Airida [17]

For Pascal's law, the pressure is transmitted with equal intensity to every part of the fluid:
$p_1 = p_2$
which becomes
$\frac{F_1}{A_1}= \frac{F_2}{A_2}$
where
$F_1=3600 N$ is the force on the first piston
$A_1=0.030 m^2$ is the area of the first piston
$F_2$ is the force on the second piston
$A_2=0.015 m^2$ is the area of the second piston

If we rearrange the equation and we use these data, we can find the intensity of the force on the second piston:
$F_2=F_1 \frac{A_2}{A_1}=(3600 N) \frac{0.015 m^2}{0.030 m^2}= 1800 N$

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3 years ago

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How does a sound wave form

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Sound waves travel through the air, liquids, and solids. they transfer energy from the source of the sound to its surroundings

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3 years ago

A certain spring exerts a nonlinear force given by F(x) = −60x − 18x2 , where x is in meters and F is in newtons. A 0.90-kg bloc

tia_tia [17]

Answer:

a) W = 6.75 J and b) v = 3.87 m / s

Explanation:

a) In the problem the force is nonlinear and they ask us for work, so we must use it's definition

W = ∫ F. dx

Bold indicates vectors. In a spring the force is applied in the direction of movement, whereby the scalar product is reduced to the ordinary product

W = ∫ F dx

We replace and integrate

W = ∫ (-60 x - 18 x²) dx

W = -60 x²/2 -18 x³/3

Let's evaluate between the integration limits, lower W = 0 for x = -0.50 m, to the upper limit W = W for x = 0 m

W = -30 [0- (-0.50) 2] -6 [0 - (- 0.50) 3]

W = 7.5 - 0.75

W = 6.75 J

b) Work is equal to the variation of kinetic energy

W = ΔK

W = ΔK = ½ m v² -0

v =√ 2W/m

v = √(2 6.75/ 0.90)

v = 3.87 m / s

8 0

3 years ago