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Helga [31]
3 years ago
15

How much work is required to lift an object with a mass of 5.0 kilograms to a height of 3.5 meters?

Physics
2 answers:
KATRIN_1 [288]3 years ago
7 0
Work is Force multiplied by distance: work = Fd

work = mgh

work = (5.0 kg)(9.8 m/s)(3.5 m)

work = 171.5 joules or for two sigfigs, 170 joules
Veronika [31]3 years ago
6 0
I believe it would be 172 J (or 1.72 · 10² J)
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jek_recluse [69]

Explanation:

Given that,

Force, F=((-8i)+6j)\ N

Position of the particle, r=(3i+4j)\ m

(a) The toque on a particle about the origin is given by :

\tau=F\times r

\tau=((-8i)+6j) \times (3i+4j)

Taking the cross product of above two vectors, we get the value of torque as :

\tau=(0+0-50k)\ N-m

(b) Let \theta is the angle between r and F. The angle between two vectors is given by :

cos\theta=\dfrac{r.F}{|r|.|F|}

cos\theta=\dfrac{(3i+4j).((-8i)+6j)}{(\sqrt{3^2+4^2} ).(\sqrt{8^2+6^2}) }

cos\theta=\dfrac{0}{50}

\theta=90^{\circ}

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3 years ago
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3 years ago
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A boy of mass 43.2 kg runs and jumps onto a stationary skateboard.The boy lands on the skateboard with a horizontal velocity of
kirza4 [7]

Answer:

= 3.87 m/s

Explanation:

Since momentum is conserved, momentum right before and after the boy jumps onto the skateboard should be the same.

Initial momentum = momentum of boy = 43.2 x 4.10 = 177 kg·m/s

Final momentum = momentum of boy and skateboard = (43.2 + 2.50) x v = 45.7v

177 = 45.7v

v = 177 / 45.7 = 3.87 m/s

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4 0
3 years ago
It is desired that 7.7 mc of charge be stored on each plate of a 5.3 mf capacitor. what potential difference is required between
antoniya [11.8K]
In physics, the elements in a circuit could involve a resistor-capacitor, resistor-inductor, or just solely their own type of circuit. For a resistor-capacitor or RC circuit, the potential difference or voltage induced between the parallel plates of a capacitor is related by this equation:

Q = C × V, where

Q is charge in Coulombs
C is the capacitance in Faradays
V is the voltage in volts

Substituting the values, 

7.7×10⁻³ C = 5.3×10⁻³ F * V
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8 0
3 years ago
A football player punts the ball from the ground at a 65.0° angle above the horizontal. If the ball stays in the air for a total
maw [93]

Answer: V_{0,x } = 297.7 \frac{m}{s}\\V_{0,y } = 637 \frac{m}{s}

Explanation:

Hi!

We define the point (0,0) as the intial position of the ball. The initial velocity is (V_{0,x}, V_{0,y})

The motion of the ball in the horizontal direction (x) has constant velocity, because there is no force in that direction. :

x(t) = V_{0,x}t

In the vertical direction (y), there is the downward acceleration g of gravity:

y(t) = -gt^2 + V_{0,y}t

(note the minus sign of acceleration, because it points in the negative y-direction)

When the ball hits the ground, at t = 65s,  y(t = 65 s) = 0. We use this to find the value of the initial vertical velocity:

0 = t(-gt + V_{0,y})\\V_{0,y} = gt = 9.8 \frac{m}{s^2} 65 s = 637 \frac{m}{s}

We used that g = 9.8 m/s²

To find the horizonttal component we use the angle:

\tan(65\º) = \frac{V_{y,0}}{V_{x,0}} = 2.14\\V_{x,0} = 297.7\frac{m}{s}

5 0
3 years ago
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