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Lelu [443]
3 years ago
8

Two masses are connected by a string which passes over a pulley with negligible mass and friction. One mass hangs vertically and

one mass slides on a horizontal surface. The horizontal surface has a coefficient of kinetic friction of 0.200. The vertically hanging mass is 3.00 kg and the mass on the horizontal surface is 3.00 kg. The magnitude of the acceleration of the vertically hanging mass is (the initial velocity of the horizontal mass is to the right)
Physics
1 answer:
monitta3 years ago
3 0

Answer:

a=2,5m/s^2

Explanation:

From the question we are told that:

Coefficient of kinetic friction \mu= 0.200

Vertical Mass M_v=3kg

Horizontal mass M_h=3.00kg  

Generally the equation for kinetic force F_k is mathematically given by

F_k=\mu N\\F_k=0.2*3\\F_k=0.6

Generally the equation for T is mathematically given by

For M_v=3kg3g-T=3a

For M_h=3kg

T=M_v V+F_k\\T=3.0a+0.6

Therefore substituting

3-3a-0.6=3a\\2.4g=6a

a=2,5m/s^2

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lora16 [44]

Answer:

gold, silver, copper, iron, lead, tin, mercury, sulfur, and carbon

Explanation:

gold, silver, copper, iron, lead, tin, mercury, sulfur, and carbon

-  (These are the most longest elements that have been around very long )

7 0
2 years ago
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The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given b
Mnenie [13.5K]

Explanation:

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :

Q=t^3-2t^2+4t+4

We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :

I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(t^3-2t^2+4t+4)}{dt}\\\\I=3t^2-4t+4

At t = 1 s,

Current,

I=3(1)^2-4(1)+4\\\\I=3\ A

So, the current at t = 1 s is 3 A.

For lowest current,

\dfrac{dI}{dt}=0\\\\\dfrac{d(3t^2-4t+4)}{dt}=0\\\\6t-4=0\\\\t=0.67\ s

Hence, this is the required solution.

7 0
3 years ago
A bird flew 16 km west in 5 hours , then flew 20 km east in 6 hours . What was the birds velocity?
LenaWriter [7]
During that period of time, the bird's displacement was 4 km east. So its velocity was (4km east)/(11hrs). That's 0.36 km/hour east. (rounded)
5 0
2 years ago
A spring is hung from the ceiling. A 2.0-kg mass suspended hung from the spring extends it by 6.0 cm. A downward external force
Stolb23 [73]

The work done by force on a spring hung from the ceiling will be 1.67 J

Any two things with mass are drawn together by the gravitational pull. We refer to the gravitational force as attractive because it consistently seeks to draw masses together rather than pushing them apart.

Given that a spring is hung from the ceiling with a 2.0-kg mass suspended hung from the spring extends it by 6.0 cm and a downward external force applied to the mass extends the spring an additional 10 cm.

We need to find the work done by the force

Given mass is of 2 kg

So let,

F = 2 kg

x = 0.1 m

Stiffness of spring = k = F/x

k = 20/0.006 = 333 n/m

Now the formula to find the work done by force will be as follow:

Workdone = W = 0.5kx²

W = 0.5 x 333 x 0.1²

W = 1.67 J

Hence the work done by force on a spring hung from the ceiling will be 1.67 J

Learn more about force here:

brainly.com/question/12970081

#SPJ4

8 0
1 year ago
A scooter is traveling at a constant speed v when it encounters a circular hill of radius r = 480 m. The driver and scooter toge
Grace [21]

Answer:

68.585m/sec , 779.1 N

Explanation:

To feel weightless, centripetal acceleration must equal g (9.8m/sec^2). The accelerations then cancel.

From centripetal motion.

F =( mv^2)/2

But since we are dealing with weightlessness

r = 480m

g = 9.8m/s^2

M also cancels, so forget M.

V^2 = Fr

V = √ Fr

V =√ (9.8 x 480) = 4704

= 68.585m/sec.

b) Centripetal acceleration = (v^2/2r) = (68.585^2/960) = 4704/960

= 4.9m/sec^2.

Weight (force) = (mass x acceleration) = 159kg x (g - 4.9)

159kg × ( 9.8-4.9)

159kg × 4.9

= 779.1N

6 0
3 years ago
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