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CaHeK987 [17]
3 years ago
15

Person is lifting a 250 N dumbbell. The weight is 30 cm from the pivot point of the elbow. What force must be exerted five from

the elbow to lift the weight? Assume everything is perpendicular.

Physics
1 answer:
qwelly [4]3 years ago
4 0
Refer to the diagram shown below.

The force, F, is applied at 5 cm from the elbow.

For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N

Answer: 1500 N

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Assuming no air resistance, all projectiles have:
Kitty [74]

Answer: Option C

C) accelerated vertical motion and constant horizontal motion.

Explanation:

If there is no air resistance then during the projectile movement the only force that causes an acceleration is the gravitational force.

We know that this force produces an acceleration of 9.8 m / s ^ 2 in the projectile.

As the gravitational force attracts the object towards the earth, then the acceleration that this force produces is always in the vertical direction. In the horizontal direction the object is not accelerated (because there is no air resistance).

Therefore the correct answer is option C.

 "accelerated vertical motion and constant horizontal motion".

3 0
3 years ago
In an experiment,the group that is exposed to the variable to be tested is called
bagirrra123 [75]

Answer:

The group that is exposed to the variable to be tested is called as experimental group.

Explanation:

An experiment consists of two kinds of group. They are experimental group and control group. The working of control group and experimental group are mostly same except that the variables will be changed in experimental group, while it is kept constant in control group. So at the end of experiments, the results from control group and experimental group are compared and the final outcome has been derived. So the given statement where the group is exposed to the variable to be tested or the independent variable is called as experimental group.

4 0
3 years ago
Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance
PSYCHO15rus [73]

Answer:

No she cannot.

Explanation:

Let v_h be the horizontal component of the ball velocity when it's kicked, assume no air resistance, this is a constant. Also let v_v be the vertical component of the ball velocity, which is affected by gravity after it's kicked.

The time it takes to travel 95m accross the field is

t = 95 / v_h or v_h = 95/t

t is also the time it takes to travel up, and the fall down to the ground, which ultimately stops the motion. So the vertical displacement after time t is 0

s = v_vt + gt^2/2= 0

where g = -9.8m/s2 in the opposite direction with v_v

v_vt - 4.9t^2 = 0

v_vt = 4.9t^2

v_v = 4.9t

Since the total velocity that the goal keeper can give the ball is 30m/s

v = v_v^2 + v_h^2 = 30^2 = 900

(4.9t)^2 + \left(\frac{95}{t})^2 = 900

24.01t^2 + \frac{9025}{t^2} = 900

Let substitute x = t^2 > 0

24.01 x + \frac{9025}{x} = 900

We can multiply both sides by x

24.01 x^2 + 9025 = 900x

24.01x^2 - 900x + 9025 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{900\pm \sqrt{(-900)^2 - 4*(24.01)*(9025)}}{2*(24.01)}

As (-900)^2 - 4*24.01*9025 = -56761 < 0

The solution for this quadratic equation is indefinite

So it's not possible for the goal keeper to do this.

6 0
3 years ago
Suppose a skydiver (mass = 80kg) is falling toward the Earth. Calculate the skydiver’s gravitational potential energy at a point
Vlad [161]
Ep is gravitational potential energy
m is mass (kg)
g is gravitational field strength (N/kg)
h is height (m)

Ep= mgh
= 80kg*9.8N/kg*100m
= 78 400 J
4 0
3 years ago
1.) Using Ohm’s law, explain how voltage changes in relation to current, assuming that resistance remains constant.
Harrizon [31]
1.) Ohm's law is understood as I = V/R. Given that resistance is constant, then voltage changes directly proportional to current.

2.) The more current that passes through a lightbulb, the brighter it glows. The higher the current, the higher the power, where power determines the brightness of a bulb.

3.) A bulb has a specific limit to how much power (Watts) it can handle. Going over the limit would cause the bulb to burn out.

4.) When a bulb burns out, no current will be able to pass through the filament.
4 0
3 years ago
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