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lara [203]
3 years ago
7

Calculate the force it would take to accelerate a 50 kg bike at a rate of 3 m/s2.

Physics
1 answer:
Ede4ka [16]3 years ago
4 0

ummmm it might be 300... i used a calculator

sorry if it is wrong

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What are the average velocity and average acceleration of the tip of a 2.4 cm long hour hand of a clock?
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A circle has a revolution of 360°. Since there are 12 hour markings, each hour interval has an angle of 30°. In radians, that would be equal to π/6 radians. So, in every 1 hour that passes, it covers π/6 of an angle. So, the angular velocity denoted as ω is π/6 ÷ 1 hour = π/6 rad/h. We can compute the average linear velocity, v, from the relationship:

v = rω, where r is the radius of the circle which is the length of the hour hand
v = (2.4 cm)(π/6 rad/h)
v = 1.257 cm/hour

Therefore, the average velocity is 1.257 cm per hour.

For the average acceleration, it is equal to zero. The hands of the clock move at a constant velocity. Since acceleration is the change of velocity per unit time, there is no change of velocity because it's constant. That's why it is zero.

8 0
3 years ago
Andre, whose mass is 77.1 kg, sits on a diving board above a dunk tank. If he is sitting 0.90 m above the water, what is his tot
Ganezh [65]

Answer:

680 J

Explanation:

Mechanical energy = potential energy + kinetic energy

ME = PE + KE

ME = mgh + ½ mv²

ME = (77.1 kg) (9.8 m/s²) (0.90 m) + ½ (77.1 kg) (0 m/s)²

ME = 680 J

8 0
3 years ago
Read 2 more answers
A particle with charge 6 mC moving in a region where only electric forces act on it has a kinetic energy of 1.9000000000000001 J
Vesna [10]

Answer:

16.9000000000000001 J

Explanation:

From the given information:

Let the initial kinetic energy from point A be K_A = 1.9000000000000001 J

and the final kinetic energy from point B be K_B = ???

The charge particle Q = 6 mC = 6 × 10⁻³ C

The change in the electric potential from point B to A;

i.e. V_B - V_A = -2.5 × 10³ V

According to the work-energy theorem:

-Q × ΔV = ΔK

-Q \times ( V_B - V_A) = (K_B - K_A)

-(6\times 10^{-3}\ C) \times ( -2.5 \times 10^3) = (K_B - 1.9000000000000001 \ J)

15 = (K_B - 1.9000000000000001 \ J)

K_B = 15+ 1.9000000000000001 \ J

\mathbf{K_B =1 6.9000000000000001 \ J}

3 0
3 years ago
studies show that the ocean surface temperature is increasing which of the following is a possible consequences
katen-ka-za [31]
One possible consequence is that the warmer temperature cause the polar ice to melt even faster
6 0
3 years ago
An electron in a hydrogen atom undergoes a transition from the n = 3 level to the n = 6 level. To accomplish this, energy, in th
Harlamova29_29 [7]

Answer:1.816\times 10^{-19} J

Explanation:

Given

E=\frac{hc}{\lambda }

E=2.18\times 10^{-18}(\frac{1}{n_1^2}-\frac{1}{n_2^2})

where h=Planck constant

c=speed of light

E=2.18\times 10^{-18}(\frac{1}{3^2}-\frac{1}{6^2})

E=2.18\times 10^{-18}\times \frac{1}{12}

E=1.816\times 10^{-19} J

7 0
3 years ago
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