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sdas [7]
2 years ago
5

Mike stands on a scale in an elevator. If the elevator is accelerating upwards with 4.9 m/s2, the scale reading is ____ times Mi

ke's weight.
Physics
1 answer:
Bas_tet [7]2 years ago
7 0

Answer:

F - M a      force exerted by scales on student

M a = M (9.8 + 4.9) m/s^2      upwards chosen as positive

a = 1.5 g        net acceleration of student  due to force of scales

W =M g       weight of student   (actual weight)

Wapp = M 1.5 * g      apparent weight (on scales) of student

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Which has more kinetic energy, a 4.0 kg bowling ball moving at 1.0 m/s or a 1.0 kg
jolli1 [7]

Answer:

The kinetic energy of bocce ball is more.

Explanation:

Given that,

Mass of a bowling ball, m₁ = 4 kg

Speed of the bowling ball, v₁ = 1 m/s

Mass of bocce ball, m₂ = 1 kg

Speed of bocce ball, v₂ = 4 m/s

We need to say which has more kinetic energy.

The kinetic energy of an object is given by :

E=\dfrac{1}{2}mv^2

Kinetic energy of the bowling ball,

E_1=\dfrac{1}{2}m_1v_1^2\\\\E_1=\dfrac{1}{2}\times 4\times (1)^2\\\\E_1=2\ J

The kinetic energy of the bocce ball,

E_2=\dfrac{1}{2}m_2v_2^2\\\\E_2=\dfrac{1}{2}\times 1\times (4)^2\\\\E_2=8\ J

So, the kinetic energy of bocce ball is more than that of bowling ball.

5 0
3 years ago
Review Conceptual Example 6 as background for this problem. A car is traveling to the left, which is the negative direction. The
DiKsa [7]

Answer:

(a) 1.21 m/s² (b) 1.75 m/s²

Explanation:

The initial speed of the car, u = 17.8 m/s

Case 1.

Final speed of the car, v = 23.5 m/s

Time, t = 4.68-s

Acceleration = rate of change of velocity

a=\dfrac{23.5 -17.8 }{4.68}\\\\a=1.21\ m/s^2

Case 2.

Final speed of the car, v = 15.3 m/s

a=\dfrac{23.5 -15.3}{4.68}\\\\a=1.75\ m/s^2

Hence, this is the required solution.

3 0
2 years ago
An astronaut exploring a distant solar system lands on an unnamed planet with a radius of 2530 km. When the astronaut jumps upwa
Natali [406]

Answer:

1.38*10^18 kg

Explanation:

According to the Newton's law of universal gravitation:

F=G*\frac{m_a*m_p}{r^2}

where:

G= Gravitational constant (6.674×10−11 N · (m/kg)2)

ma= mass of the astronaut

mp= mass of the planet

F=m_a.a\\(v_f )^2=(v_o)^2+2.a.\Delta y\\\\a=\frac{(v_f)^2-(v_o)^2}{2.\Delta y}\\\\a=\frac{(0)^2-(4.29m/s)^2}{2.0.64m}=14.38m/s^2\\\\F=m_a*14.38m/s^2

so:

m_a*14.38m/s^2=(6.674*10^{-11}N.(m/kg)^2)*\frac{m_a.m_p}{(2.530*10^3m)^2}\\m_p=\frac{14.38m/s^2(2.530*10^3m)^2}{(6.674*10^{-11}N.(m/kg)^2)}\\\\m_p=1.38*10^{18}kg

7 0
3 years ago
You kick a ball with a speed of 14 m/s at an angle of 51°. How far away does the ball land?
In-s [12.5K]
-- The vertical component of the ball's velocity is 14 sin(<span>51°) = 10.88 m/s

-- The acceleration of gravity is 9.8 m/s².

-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for the
same amount of time before it hits the ground.

-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds
==================================

-- The horizontal component of the ball's velocity is  14 cos(</span><span>51°) = 8.81 m/s

-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = <em><u>19.56 meters</u></em>
before it hits the ground.


As usual when we're discussing this stuff, we completely ignore air resistance.
</span>
4 0
3 years ago
Read 2 more answers
A voltage source provides ___________ required for electric current.
Gnoma [55]
Two terminal device which can maintain a fixed voltage. Hope this help I need more detail in your question this is all I can provide. :)
8 0
3 years ago
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