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Alecsey [184]
3 years ago
8

A person with a mass of 72 kg is riding a bicycle is accelerating at a rate of 5m/s on a horizontal surface. What is the weight

force the person exerts on the bicycle seat
Physics
1 answer:
nikklg [1K]3 years ago
7 0

Answer:

w = 706.32 [N]

Explanation:

The force due to gravitational acceleration can be calculated by means of the product of mass by gravitational acceleration.

w = m*g

where:

w = weight [N] (units of Newtons)

m = mass = 72 [kg]

g = gravity acceleration = 9.81 [m/s²]

Then we have:

w = 72*9.81\\w = 706.32 [N]

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A guitar string sends out waves that hit your ear. in 2 seconds, 76 wave cycles hit your ear
Novosadov [1.4K]

Answer:

a

Explanation:

8 0
3 years ago
Two charges are located in the x x – y y plane. If q 1 = − 2.75 nC q1=−2.75 nC and is located at ( x = 0.00 m , y = 0.600 m ) (x
ludmilkaskok [199]

Answer:

\vec{E}=(-15.78\hat{i}+63.81\hat{j})\frac{N}{C}

Explanation:

In this case we have to work with vectors. Firs of all we have to compute the angles between x axis and the r vector (which points the charges):

\theta_1=tan^{-1}(\frac{0.6m}{0m})=90\° \\\\\theta_2=tan^{-1}(\frac{0.4m}{1.3m})=17.10\°

the electric field has two components Ex and Ey. By considering the sign of the charges we obtain that:

\vec{E} = E_x \hat{i}+E_y\hat{j}\\\\\vec{E}=(E_1cos\theta_1-E_2cos\theta_2)\hat{i}+(E_1sin\theta_1-E_2sin\theta_2)\hat{j}\\\\E_1=k\frac{q}{r_1^2}=(8.99*10^{9}\frac{N}{m^2C^2})\frac{2.75*10^{-9}C}{(0.6m)^2}=68.67\frac{N}{C}\\\\E_2=k\frac{q}{r_2^2}=(8.99*10^{9}\frac{N}{m^2C^2})\frac{3.40*10^{-9}C}{((1.3m)^2+(0.4m)^2)}=16.52\frac{N}{C}

Hence, by replacing E1 and E2 we obtain:

\vec{E}=[(68.67N/C)cos(90\°)-(16.52N/C)cos(17.10\°)]\hat{i}+[(68.67N/C)sin(90\°)-(16.52N/C)sin(17.10\°)]\hat{j}\\\\\vec{E}=(-15.78\hat{i}+63.81\hat{j})\frac{N}{C}

hope this helps!!

3 0
3 years ago
What do you think happened to the energy you transferred to the notebook when you pushed it?
xeze [42]

Answer:

It got transferred to kinetic energy

Explanation:

8 0
3 years ago
Read 2 more answers
What is the value of Ax?
dusya [7]
Ax<span>”= ot * k-> 00 A(i+2) == X:"40: hence lim </span>Ax<span>") = 0</span>
4 0
3 years ago
A layer of liquid B floats on liquid A. A ray of light begins in liquid A and undergoes total internal reflection at the interfa
amm1812

Answer:

0.737

Explanation:

n_{A} = Refractive indices of liquid A

n_{B} = Refractive indices of liquid B

n_{C} = Refractive indices of liquid C

Consider the total internal reflection at interface of liquid A and liquid B

\theta_{i} = Angle of incidence = 32.0

Using Snell's law for total internal reflection

n_{A} Sin\theta_{i} = n_{B} \\n_{B} = n_{A} Sin32

Consider the total internal reflection at interface of liquid A and liquid C

\theta_{i} = Angle of incidence = 46

Using Snell's law for total internal reflection

n_{A} Sin\theta_{i} = n_{C} \\n_{C} = n_{A} Sin46

Ratio is hence given as

Ratio = \frac{n_{B}}{n_{C}} = \frac{n_{A} Sin32}{n_{A} Sin46} = \frac{Sin32}{Sin46}\\Ratio = 0.737

7 0
3 years ago
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