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3 years ago

I need HELP ASAP please...

Physics

2 answers:

Ahat [919]3 years ago

8 0

Answer:

B. discover what interests you

Explanation:

A career decision must be based on one's interest. A career is something we have to do life long. And if it to be done life long, we must be vary careful in deciding are career. A believe we must identify our interest first. And experiment with to find the best combination of interest and future prospect. This cannot be done by the candidate itself though. A proper guidance by parents and a mentor is required

Hunter-Best [27]3 years ago

3 0

B for sure, you can’t do anything without knowing what you’re interested in!

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Say you want to make a sling by swinging a mass M of 2.3 kg in a horizontal circle of radius 0.034 m, using a string of length 0

ycow [4]

Answer:

T = 764.41 N

Explanation:

In this case the tension of the string is determined by the centripetal force. The formula to calculate the centripetal force is given by:

$F_c=m\frac{v^2}{r}$ (1)

m: mass object = 2.3 kg

r: radius of the circular orbit = 0.034 m

v: tangential speed of the object

However, it is necessary to calculate the velocity v first. To find v you use the formula for the kinetic energy:

$K=\frac{1}{2}mv^2$

You have the value of the kinetic energy (13.0 J), then, you replace the values of K and m, and solve for v^2:

$v^2=\frac{2K}{m}=\frac{2(13.0J)}{2.3kg}=11.3\frac{m^2}{s^2}$

you replace this value of v in the equation (1). Also, you replace the values of r and m:

$F_c=(2.3kg)(\frac{11.3m^2/s^2}{0.034})=764.41N$

hence, the tension in the string must be T = Fc = 764.41 N

5 0

2 years ago

Why does a third class lever cannot magnify force?

maw [93]

Explanation:

The third class lever cannot magnify our force because in third class lever the effort it between the load and the fulcrum. Also, in this type of lever no matter where the force is applied, it is always greater than the force of load. Hence, That type of lever cannot magnify our force.

5 0

2 years ago

A point charge q1 is held stationary at the origin. A second charge q2 is placed at point a, and the electric potential energy o

likoan [24]

Answer:

$U_{b}=+7.3*10^{-8}J$

Explanation:

Given data

Electric potential at point a is Ua=5.4×10⁻⁸J

q₂ moves to point b where a negative work done on it $W_{a-b}=-1.9*10^{-8}J$

Required

Electric potential energy Ub

Solution

When a particle moves from a point where the potential is Ua to a point where it is Ub the change in potential energy is equal to work done where the force exerted on the charge is conservative and work done is given by:

$W_{a-b}=U_{a}-U_{b}\\U_{b}=U_{a}-W_{a-b}$