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Gekata [30.6K]
4 years ago
14

A stack of books rests on a level frictionless surface. A force F acts on the stack, and it accelerates at 3.0 m/s2. A 1.0 kg bo

ok is then added to the stack. The same force is applied, and now the stack accelerates at 2.0 m/s2.
Physics
1 answer:
Oxana [17]4 years ago
4 0

Answer:

m1 = 2 kg

m2 = 3kg

Explanation:

The force can be getting by

F = m * a

F1 = F2

a1 = 3.0 m/s^2

a2 = 2.0 m/s^2

The force F1=F2 because the force is applied so get the a2 acceleration

m1 * a1 = m2 * a2

m2 = m1 + 1kg

m1 *(3.0 m/s^2) = m2* (2.0 m/s^2)

m1 *(3.0 m/s^2) = (m1 + 1kg) * (2.0 m/s^2)

m1*(3.0m/s^2-2.0m/s^2)=2 kg*m/s^2

Solve to find the mass

m1 m/s^2= 2 kg*m/s^2

m1 = 2 kg

m2 = 3kg

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In a particular beam of radiation, which is traveling in a vacuum, the amounts of energy per second at an ultraviolet wavelength
Darina [25.2K]

Answer: d)

Explanation: In order to justify the answer we have to consider that the energy of photons directely depent on the frequency so the energy is inverselly dependent of the wavelegth.

If both beams have the same power, this means Energy/time so the number of photons per second must be different. As consequence a) is wrong as  b) since it is not posible since UV photon  have more energy that IR photons. c) It is no necessary know the frequency since the wavelength is related in the form:

c=λν  c is the speed of light, λ the wavelegth and ν the frequency.

d) Certainly will be more more IR photons than UV photons to get the same beam power.

8 0
3 years ago
After a check up, a person now has a far point of 100 cm, but with good near point vision. He needs to wear a new pair of correc
lakkis [162]

Answer:

so his far point according to this pair of glass is 200 cm

Explanation:

power of old pair of corrective glasses is given as

P = -0.5 dioptre

now we have

f = \frac{1}{P}

f = -2 m

f = -200 cm

now we know that for normal vision the maximum distance of vision is for infinite distance

so by lens formula we have

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

\frac{1}{d_i} + 0 = \frac{1}{200}

d_i = 200 cm

so his far point according to this pair of glass is 200 cm

7 0
3 years ago
A projectile of mass 1.800 kg approaches a stationary target body at 4.800 m/s. The projectile is deflected through an angle of
Artyom0805 [142]

Answer:

P = 5.22 Kg.m/s

Explanation:

given,

mass of the projectile = 1.8 Kg

speed of the target = 4.8 m/s

angle of deflection = 60°

Speed after collision = 2.9 m/s

magnitude of momentum after collision = ?

initial momentum of the body = m x v

                                                  = 1.8 x 4.8 = 8.64 kg.m/s

final momentum after collision

momentum along x-direction

P_x = m v cos θ

P_x = 1.8 x 2.9 x  cos 60°

P_x = 2.61 kg.m/s

momentum along y-direction

P_y = m v sin θ

P_y = 1.8 x 2.9 x  sin 60°

P_y = 4.52 kg.m/s

net momentum of the body

P = \sqrt{P_x^2 + P_y^2}

P = \sqrt{4.52^2 + 2.61^2}

 P = 5.22 Kg.m/s

momentum magnitude after collision is equal to P = 5.22 Kg.m/s

5 0
3 years ago
A basketball star covers 3.05 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled pr
rosijanka [135]
<span><span>anonymous </span> 4 years ago</span>Any time you are mixing distance and acceleration a good equation to use is <span>ΔY=<span>V<span>iy</span></span>t+1/2a<span>t2</span></span> I would split this into two segments - the rise and the fall. For the fall, Vi = 0 since the player is at the peak of his arc and delta-Y is from 1.95 to 0.890. For the upward part of the motion the initial velocity is unknown and the final velocity is zero, but motion is symetrical - it takes the same amount of time to go up as it does to go down. Physiscists often use the trick "I'm going to solve a different problem, that I know will give me the same answer as the one I was actually asked.) So for the first half you could also use Vi = 0 and a downward delta-Y to solve for the time. Add the two times together for the total. The alternative is to calculate the initial and final velocity so that you have more information to work with.
5 0
3 years ago
Read 2 more answers
An object is traveling on a circle with a radius of 6 feet. If in 80 seconds a central angle of 9/4 radians is swept out, then f
trapecia [35]

Answer:

The angular speed of the object is 0.0281 rad/s

The linear speed of the object is 0.169 ft/s

Explanation:

Given;

radius of the circle, r = 6 ft

time of motion of the object around the circle, t = 80 s

central angle formed by the object during the motion, θ = 9/4 rad = 2.25 rad

The angular speed of the object is calculated as;

\omega = \frac{\theta }{t} = \frac{2.25 \ rad}{80 \ s} = 0.0281 \ rad/s

The linear speed of the object is calculated as;

v = ωr

v = 0.0281 rad/s   x    6ft

v = 0.169 ft/s

8 0
3 years ago
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