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AlexFokin [52]
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? *
Physics
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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Zinaida [17]
Hello,


I'd be glad to help.

Momentum  =  (mass) times (speed)

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                                           =         1,350 kg-m/s  .    

Hope this helps
5 0
3 years ago
physics a flower pot falls from a windowsill 25.0m above the sidewalk. how much time does a passerby on the sidewalk below have
Usimov [2.4K]
v_f^2 = v_0^2 + 2*a*d
Since the flower pot is dropped, it has an initial velocity of zero. Also the flower pot accelerates due to gravity.

so

v_f = \sqrt{2*g*d}=\sqrt{2*9.81*25.0}
8 0
3 years ago
You wish to produce an emf of 41.0 mV using an inductor whose inductance is 13.0 H. You start with a current of 1.50 mA through
Molodets [167]

Answer:

The current through the inductor at the end of 2.60s is 9.7 mA.

Explanation:

Given;

emf of the inductor, V = 41.0 mV

inductance of the inductor, L = 13 H

initial current in the inductor, I₀ = 1.5 mA

change in time, Δt = 2.6 s

The emf of the inductor is given by;

V = L\frac{di}{dt} \\\\V = \frac{L(I_1-I_o)}{dt} \\\\L(I_1-I_o) = V*dt\\\\I_1-I_o = \frac{V*dt}{L}\\\\I_1 =  \frac{V*dt}{L} + I_o\\\\I_1 = \frac{41*10^{-3}*2.6}{13} +1.5*10^{-3}\\\\I_1 = 8.2*10^{-3} + 1.5*10^{-3}\\\\I_1 = 9.7 *10^{-3} \ A\\\\ I_1 = 9.7 \ mA

Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.

6 0
3 years ago
Unit 5 lesson 7 physical science 12 question
Kryger [21]
I just took it 100% 11/11
1.D
2.A
3.A
4.A
5.B
6.C
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9A
10B
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8 0
3 years ago
Where would you find convergent and divergent plate boundaries relative to
Sergio [31]

Answer:

When two tectonic plates meet, we get a “plate boundary.” There are three major types of plate boundaries, each associated with the formation of a variety of geologic features.

Explanation:

8 0
2 years ago
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