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larisa [96]
3 years ago
14

A worker pushes a crate horizontally across a warehouse floor with a force of 245 N at an angle of 55 degrees below the horizont

al.
How much of the worker’s force is not used to move the crate?

a. 200.69 N
b. 419.95 N
c. 357.71 N
d. 191.32 N
Physics
2 answers:
aev [14]3 years ago
6 0

Answer:

option A

Explanation:

given,

For exerted by the worker = 245 N

angle made with horizontal = 55°

we need to calculate Force which is not used to move the crate = ?

Movement of crate is due to the horizontal component of the force.

Crate will not move due to vertical force acting on the it.

F_y = F sin \theta

F_y = 245\times sin 55^0

F_y =200.69

hence, worker's force not used to move the crate is equal to 200.69

The correct answer is option A

ivanzaharov [21]3 years ago
4 0

Answer:

The worker’s force is 200.69 N.

(a) is correct option.

Explanation:

Given that,

Force = 245 N

Angle = 55°

According to component of force

Horizontal component of force is in direction of displacement which is used to move the crate.

Vertical component of force is not used to move the crate.

We need to calculate the worker’s force which is not used to move the crate

Using y- component of force

F_{y}=F\sin\theta

Put the value into the formula

F_{y}=245\times\sin55

F_{y}=200.69\ N

Hence, The worker’s force is 200.69 N.

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Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

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Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

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F = \frac{k|q_1||q_2|}{r^2}

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F is the force of attraction between the two charges

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k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N

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