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Olenka [21]
2 years ago
7

The apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ?

use negative is the acceleration vector is pointing downwards
​
Physics
1 answer:
Yuki888 [10]2 years ago
4 0

Answer:

-.457 m/s^2

Explanation:

Actual weight =   60 .3 (9.81) = 591.54 N

Accel of lift changes this to    60.3 ( 9.81 - L)     where L - accel of lift

                                           60.3 ( 9.81 - L ) = 564

                                               solve for L = .457 m/s^2  DOWNWARD

                                                        so L = - .457 m/s^2

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Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t
salantis [7]

Explanation:

Mass of bumper cars, m_1=m_2=110\ kg

Initial speed of car A, u_1=8\ m/s

Initial speed of car Z, u_2=-10\ m/s

Final speed of car A after the collision, v_1=-10\ m/s

We need to find the velocity of car Z after the collision. Let it is equal to v_2. Using the conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

110\times 8+110\times (-10)=110\times (-10)+110v_2

v_2=\dfrac{-1320}{110}\ m/s

v_2=-12\ m/s

So, the velocity of car Z after the collision is (-12 m/s). Hence, this is the required solution.

5 0
3 years ago
The figure below shows two forces A=15 N and B=5 N acting on an object. What is the
alexandr1967 [171]

Answer:

18.03 N

Explanation:

From the fiqure below,

Using parallelogram law of vector

R² = 15²+5²-2×5×15cos(180-60)

R² = 225+25-150cos120°

R² = 250-150(-0.5)

R² = 250+75

R² = 325

R = √325

R = 18.03 N

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7 0
3 years ago
A cylinder of mass 14.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed o
aev [14]

Answer:

a) 567J

b) 283.5J

c)850.5J

Explanation:

The expression for the translational kinetic energy is,

E_r = \frac{1}{2} mv^2

Substitute,

14kg for m

9m/s for v

E_r = \frac{1}{2} (14) (9)^2\\= 567J

The translational kinetic energy of the center of mass is 567J

(B)

The expression for the rotational kinetic energy is,

E_R = \frac{1}{2} Iw^2

The expression for the moment of inertia of the cylinder is,

I = \frac{1}{2} mr^2

The expression for angular velocity is,

w = \frac{v}{r}

substitute

1/2mr² for I

and vr for w

in equation for rotational kinetic energy as follows:

E_R = (\frac{1}{2}) (\frac{1}{2} mr^2)(\frac{v}{r} )^2

= \frac{mv^2}{4}

E_R = \frac{14 \times 9^2 }{4} \\\\= 283.5J

The rotational kinetic energy of the center of mass is 283.5J

(c)

The expression for the total energy is,

E = E_r + E_R\\\\

substitute 567J for E(r) and 283.5J for E(R)

E = 567J + 283.5\\= 850.5J

The total energy of the cylinder is 850.5J

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This is A.) lake. A river is a small amount of water that isn't always fresh water. A stream is too small. And an ocean is made of salt water.
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