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seraphim [82]
3 years ago
9

Ice-skater slides toward a sled sitting on the ice and hits it. The skater exerts a 12.6 N force on the sled at an angle of 15.3

° below the horizontal. The sled then moves 15.4 m forward. How much work did the skater do on the sled? Assume friction is negligible.
Knowns Unknown

F = ____________ d = _____________ W = ?

θ = ____________

Solve for the Unknown

a. Write the equation for work
b. Insert known quantities
Physics
1 answer:
RideAnS [48]3 years ago
5 0

Answer:

Expression of work done is

W = Fd cos\theta

Work done to move the sled is given as 187.2 J

Explanation:

As we know that the formula of work done is given as

W = Fd cos\theta

here we know that

F = 12.6 N

d = 15.4 m

\theta = 15.3 degree

so we will have

W = 12.6 \times 15.4 cos15.3

W = 187.2 J

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Hope this helps!

7 0
3 years ago
A (20*20) cm² loop has a resistance of 0.10 Ω. A magnetic field perpendicular to the loop is B = 4t - 2t², where B is in tesla a
Ilya [14]

Answer with Explanation:

We are given that

Area of loop=(20\times 20) cm^2=400\times 10^{-4} m^2

1 cm^2=10^{-4} m^2

Resistance, R=0.1\Omega

B=4t-2t^2

We know that magnetic flux

\phi=BA

Emf ,E=\mid \frac{d\phi}{dt}\mid =\mid\frac{d(BA}{dt}\mid =\mid A\frac{dB}{dt}=400\times 10^{-4}\times \frac{4t-2t^2}{dt}\mid =\mid400\times 10^{-4}\times(4-4t)\mid

Current, I=\frac{E}{R}

Current, I=\frac{\mid 400\times 10^{-4}(4-4t)\mid }{0.1}=1.6\mid (1-t)\mid

Substitute t=0 s

Then, I=1.6\mid (1-0)\mid=1.6 A

Substitute t=1 s

Then, I=1.6\mid (1-1)\mid=0

Substitute

t=2 s

Current, I=1.6\mid(1-2)\mid=1.6 A

8 0
3 years ago
Read 2 more answers
If 34.7 g of O2 reacts with iron to form 79.34 g of iron oxide, how much iron was used in the reaction?
zhuklara [117]

Answer: B. 44.64 g

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Mass of reactants = mass of iron + mass of oxygen = mass of iron + 34.7 g

Mass of product = mass of iron oxide = 79.34 g

As Mass of reactants = Mass of product

mass of iron + 34.7 g = 79.34 g

mass of iron = 44.64 g

Thus 44.64 g of iron was used in the reaction

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