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2 years ago

Determine the value of the force exerted by the surface (normal force) on a

Physics

1 answer:

Advocard [28]2 years ago

8 0

Answer:

53.5 N

Explanation:

Vertical component of the F force 50 sin30 = 25 N upward

force of gravity = m g = 8 * 9.81 =78.5 N Downward

NET downward force by block on table = net upward force exerted by table = 78.5 -25 =53.5 N

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Convert 27,549 into scientific notation

dezoksy [38]

2.7549 x 10^4 is the answer I hope this helped u

7 0

3 years ago

A student decides to spend spring break by driving 50 miles due east, then 50 miles 30 degrees south of east, then 50 miles 30 d

PolarNik [594]

Answer:

a. 600 ml, 12

Explanation:

The movement described in the question exhibits that of a polygon. Exhibiting a constant distance and angle with only varying direction until the starting point is reached.

The sum of exterior angles of a polygon = 360 degrees.

Exterior angle of a polygon = (360 ÷ number of sides)

Therefore,

Number of sides = 360 ÷ exterior angle

Exterior angle = 30 degrees

Hence,

Number of sides = 360 ÷ 30 = 12 sides

Since distance traveled of 50 miles is the same for each displacement ;

Total displacement = distance traveled * number of sides

Total displacement = 50 * 12 = 600 miles.

5 0

4 years ago

A scientific law tells us what nature will do under certain conditions.<br> True<br> False

sergij07 [2.7K]

Hello There!

This Is A "True" Statement.

<em>A scientific law tells us what nature will do under certain conditions.</em>

6 0

3 years ago

A point charge, Q1 = -4.2 μC, is located at the origin. A rod of length L = 0.35 m is located along the x-axis with the near sid

igor_vitrenko [27]

Answer:

a) attractiva, b) dF = $k \frac{Q_1 \ dQ_2}{dx}$ , c) F = $k Q_1 \frac{Q_2}{d \ (d+L)}$ , d) F = -1.09 N

Explanation:

a) q1 is negative and the charge of the bar is positive therefore the force is attractive

b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x

dF = $k \frac{Q_1 \ dQ_2}{dx}$

where k is a constant, Q₁ the charge at the origin, x the distance

c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L

∫ dF = $k \ Q_1 \int\limits^{d+L}_d {\frac{1}{x^2} } \, dQ_2$

as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density

λ = dQ₂ / dx

DQ₂ = λ dx

we substitute

F = $k \ Q_1 \lambda \int\limits^{d+L}_d \, \frac{dx}{x^2}$

F = k Q1 λ ( $-\frac{1}{x}$ )

we evaluate the integral

F = k Q₁ λ $(- \frac{1}{d+L} + \frac{1}{d} )$

F = k Q₁ λ $( \frac{L}{d \ (d+L)})$

we change the linear density by its value

λ = Q2 / L

F = $k Q_1 \frac{Q_2}{d \ (d+L)}$

d) we calculate the magnitude of F

F =9 10⁹ (-4.2 10⁻⁶) $\frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}$

F = -1.09 N

the sign indicates that the force is attractive

3 0

3 years ago

A 25 N force at 60° is required to set a crate into motion on a floor. What is the value of the static friction?

True [87]

The normal force of the force given is calculated through the equation,

Fn = F(sin θ)
where Fn is the normal force, F is the force, and θ is the angle.

Fn = (25 N)(sin 60°) = 21.65 N

The x-component of the force applied is,
Fx = (25 N)(cos 60°) = 12.5 N

The value of the coefficient of static friction is calculated through the equation,
F = μFn
μ = Fx / Fn = 12.5 N / 21.65 N = 0.577

3 0

3 years ago