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jek_recluse [69]
3 years ago
6

A waiter is carrying a tray above his head and walking at a constant velocity. If he applies a force of 5.0 newtons on the tray

and covers a distance of 10.0 meters, how much is the work being done?
A. 0 joules
B. 2 joules
C. -2 joules
D. 50 joules
E. -50 joules
Physics
1 answer:
Simora [160]3 years ago
4 0

Answer:

(A) 0 Joules

Explanation:

The force applied exists in the vertical direction only. There is no displacement in the vertical direction. (Work)=(Force in y direction) x (Displacement in y direction) = (Force) x 0 = 0 J

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A sprinter has a mass of 80kg and a kinetic energy of 4000 j. What is the sprinter's speed
JulijaS [17]
The formula of the kinetic energy is KE = 0.5*m*v^2.
Given m = 80 kg and KE = 4000 J,
4000 = 0.5*80*v^2
v^2 = 100
v = 10 m/s
4 0
3 years ago
A small boat is moving at a velocity of 1.93 m/s when it is accelerated by a river current perpendicular to the initial directio
olasank [31]

Answer:

16.405m/s

Explanation:

Given, initial velocity = u = 1.93m/s, acceleration = a = 0.750m/s2, time = t = 19.3s, final velocity = v= ?

Using the first equation of linear motion,

v = u + at

v = 1.93 + 0.750 x 19.3

v = 1.93 + 14.475

v = 16.405m/s

7 0
3 years ago
Read 2 more answers
A stone is dropped from rest from the top of a building. It takes Δt = 2.2 s for it to reach the ground.
NeTakaya

Answer:

Value of magnitude of acceleration will be 9.8m/sec^2

Explanation:

It is given that when a stone is dropped it takes 2.2 sec to reach the ground

(a) As the stone is dropped from the top of building

So its initial velocity u_i will be 0 m /sec

(b) As the stone is free falling and there is no external force applied on it so its acceleration will be equal to acceleration due to gravity

So value of magnitude of acceleration will be equal to 9.8m/sec^2

6 0
3 years ago
A series RLC circuit consists of a 52.0 Ω resistor, a 4.80 mH inductor, and a 330 nF capacitor. It is connected to an oscillator
Tju [1.3M]

Answer:

(G) 75.11 ohm

(H) 0.08 A

(I) 46.2 degree

Explanation:

R = 52 ohm

L = 4.8 m H = 4.8 x 106-3 H

C = 330 nF = 330 x 10^-9 F

Vo = 6 V

(G)

f = 5000 Hz

Let the impedance is Z.

X_{L}= 2 \pi fL = 2 \times 3.14\times 5000\times 4.8\times 10^{-3}=150.72 ohm

X_{c}= \frac{1}{2 \pi fC}=\frac{1}{2\times 3.14\times 5000\times 330\times 10^{-9}}=96.51 ohm

Z=\sqrt{R^{2}+\left ( X_{L}-X_{c} \right )^{2}}

Z=\sqrt{52^{2}+\left (150.72-96.51)^{2}}=75.11 ohm

(H) Let Io be the peak current

I_{0}=\frac{V_{0}}{Z}=\frac{6}{75.11}=0.0798 A = 0.08 A

(I) Let Ф be the phase angle

tan\phi = \frac{X_{L}-X_{C}}{R}

tan\phi =\frac{150.72-96.51}}{52}=1.0425

Ф = 46.2 degree

4 0
3 years ago
Im back and need help
const2013 [10]

Always here to help. Bring it!!!

7 0
3 years ago
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