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USPshnik [31]
4 years ago
11

What is the value of work done on an object when a 70-newton force moves it 9.0 meters in the same direction as the force?

Physics
2 answers:
ElenaW [278]4 years ago
8 0

6.3 × 102 joules on plato

marta [7]4 years ago
4 0
630 watts - 9 x 70 = 630
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Why do nations should establish a set of rules and principles for responsible lunar/moon explortions? ASAP pleaseeee :(
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3 years ago
An electric field of magnitude 2.35 V/m is oriented at an angle of 25.0° with respect to the positive z-direction. Determine the
zzz [600]

Answer:

The magnitude of the electric flux is 3.53\ N-m^2/C

Explanation:

Given that,

Electric field = 2.35 V/m

Angle = 25.0°

Area A= 1.65 m^2

We need to calculate the flux

Using formula of the magnetic flux

\phi=E\cdot A

\phi = EA\cos\theta

Where,

A = area

E = electric field

Put the value into the formula

\phi=2.35\times1.65\times\cos 25^{\circ}

\phi=2.35\times1.65\times0.91

\phi=3.53\ N-m^2/C

Hence, The magnitude of the electric flux is 3.53\ N-m^2/C

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3 years ago
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8 0
3 years ago
A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release
Svetradugi [14.3K]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done  to compress the spring initially=\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J

b.

By conservation law of energy

Initial energy of spring=Kinetic energy  of object

37.8=\frac{1}{2}(3)v^2

v^2=\frac{37.8\times 2}{3}

v=\sqrt{\frac{37.8\times 2}{3}}

v=5.02 m/s

c.Work done by friction on the incline,w_{friction}=P.E-spring \;energy

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8 0
3 years ago
A spring stretches 0.2 cm per newton of applied force. An object is suspended from the spring and a deflection of 3 cm is observ
siniylev [52]

Answer:

m=1.53kg    

Explanation:

To solve this problem we use the Hooke's Law:

F=k*\Delta x     (1)

F is the Force needed to expand or compress the spring by a distance Δx.

The spring stretches 0.2cm per Newton, in other words:

1N=k*0.2cm ⇒ k=1N/0.2cm=5N/cm  

The force applied is due to the weight

F=mg

We replace in (1):

mg=k*\Delta x  

We solve the equation for m:

m=k*\Delta x/g=5*3/9.81=1.53kg    

7 0
3 years ago
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