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3 years ago

What is the force of a 13kg ball that has been dropped and has fallen for 1 second?

Physics

1 answer:

ankoles [38]3 years ago

6 0

Answer:

127.4 newtons

Explanation:

Assuming g = 9.8:

F = ma = 13(9.8) = 127.4 N

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8. A receptacle, plug, or any other electrical device whose design limits the ability of an electrician to come in contact with

yaroslaw [1]

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4 years ago

A sharp knife cut things easily

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3 years ago

Two forces, 80N and 100N acting at an angle of 60 degrees with each other, pull on an object. What single force would replace th

KIM [24]

We use the law of Cosines, resultant force

$F =\sqrt{(F_{1})^2 + (F_{2})^2 + 2 F_{1} F_{2} cos \theta }$

Here, $(F_{1})$ and $(F_{2})$ are forces acting at angle $\theta$ with each other.

Given $F_{1} = 80 N$ , $F_{2} = 100 N$ and $\theta = 60^0$ .

Substituting these given values in above formula we get

$F =\sqrt{(80)^2 + (100)^2 + 2 \times 80 \times 100 \ cos 60^0 } = 156. 20 \ N$ .

Thus, the resultant force is 156 N.

8 0

4 years ago

A 76g ball is attached to a string that is 1.5m long. It is spun so that it completes two full rotations every second. What is t

Marat540 [252]

The centripetal acceleration = 236.63 m/s²

The force = 17.98 N

<h3>Further explanation</h3>

Given

mass = 76 g = 0.076 kg

r = 1.5 m

f = 2 rps = 2 rotation per second

Required

The centripetal acceleration

The Force tension

Solution

Centripetal force is a force acting on objects that move in a circle in the direction toward the center of the circle

$\large {\boxed {\bold {F = \frac {mv ^ 2} {R}}}$

F = centripetal force, N

m = mass, Kg

v = linear velocity, m / s

r = radius, m

The speed that is in the direction of the circle is called linear velocity

Can be formulated:

$\tt \displaysyle v = 2 \pi.r.f$

r = circle radius

f = rotation per second (RPS)

The linear velocity : 2 x 3.14 x 1.5 x 2 =18.84 m/s

The centripetal acceleration : ac = v²/R = 236.63 m/s²

The force : F = m x ac = 0.076 x 236.63 = 17.98 N

4 0

3 years ago

The solid shaft with a 20 mm radius is used to transmit the torques applied to the gears. Determine the maximum torsional shear

AysviL [449]

Answer:

τ = (7.96 x 10⁴ m⁻³)T

This is the expression for maximum allowable shear stress in terms of the maximum torque applied in Nm.

Explanation:

The maximum allowable shear stress on the solid shaft can be given by the torsional formula as follows:

τ = Tc/J

where,

τ = Maximum Allowable Shear Stress = ?

T = Maximum Torque Applied to the Shaft

c = maximum distance from center to edge = radius in this case = 20 mm = 0.02 m

J = Polar Moment of inertia = πr⁴/2 = π(0.02 m)⁴/2 = 2.51 x 10⁻⁷ m⁴

Therefore,

τ = T(0.02 m)/(2.51 x 10⁻⁷ m⁴)

<u>This is the expression for maximum allowable shear stress in terms of the maximum torque applied in Nm.</u>

3 0

3 years ago