A box took 5 minutes to move you did 2150 j of work in the process. how much power was required to move the box

The length of the minute hand of the clock is 14cm. Calculate the speed with which the tip of the minute hand moves

Molodets [167]

Speed of the tip of the minute hand=V=0.0244 cm/s

Explanation:

The angular velocity of the minute hand is given by

$\omega= \frac{2\pi}{T}$

T= time period of the minute hand=60 min=3600 s

so ω= 2 π/3600 rad/s

Now linear velocity v= r ω

r= radius of minute hand=14 cm

so v= 14 (2 π/3600)

V=0.0244 cm/s

8 0

3 years ago

A 40kg load is raised to a height of 25m. If the operation requires 1 min, find the power required

OlgaM077 [116]

Answer:

163.33 Watts

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 40 Kg

Height (h) = 25 m

Time (t) = 1 min

Power (P) =..?

Next, we shall determine the energy. This can be obtained as follow:

Mass (m) = 40 Kg

Height (h) = 25 m

Acceleration due to gravity (g) = 9.8 m/s²

Energy (E) =?

E = mgh

E = 40 × 9.8 × 255

E = 9800 J

Finally, we shall determine the power. This can be obtained as illustrated below:

Time (t) = 1 min = 60 s

Energy (E) = 9800 J

Power (P) =?

P = E/t

P = 9800 / 60

P = 163.33 Watts

Thus, the power required is 163.33 Watts

8 0

3 years ago

A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors

Mars2501 [29]

Answer:

(A) Capacitance per unit length = $4.02 \times 10^{-10}$

(B) The magnitude of charge on both conductor is $Q = 4.22 \times 10^{-19}$ C and the sign of charge on inner conductor is $+Q$ and the sign on outer conductor is $-Q$

Explanation:

Given :

Radius of inner part of conductor $(R_{1})$ = $2.7 \times 10^{-3}$ m

Radius of outer part of conductor $(R_{2})$ = $3.1 \times 10^{-3}$ m

The length of the capacitor $(l)$ = $3 \times 10^{-3}$ m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,

$C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }$

Where, $\epsilon_{o} = 8.85 \times 10^{-12}$

But we need capacitance per unit length so,

$\frac{C}{l} = \frac{2\pi\epsilon_{o} }{ln\frac{R_{2} }{R_{1} } }$

capacitance per unit length = $\frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}$

(B)

The charge on both conductors is given by,

$Q = C \Delta V$

Where, C = capacitance of cylindrical capacitor and value of $C = 12.06 \times 10^{-13}$ F, $\Delta V = 350 \times 10^{-3}$ V

∴ $Q = 4.22 \times 10^{-19}$ C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is $+Q$ and Charge on outer conductor is $-Q$ .

8 0

3 years ago

Kinetic energy defined

SVEN [57.7K]

Kinetic Energy means <span>energy that a body possesses by virtue of being in motion. I hope im not too late</span>

7 0

3 years ago

One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side

anyanavicka [17]

Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

$L1^2=L2^2=L^2=2^2+(H/2)^2$ Replacing this value in the previous equation:

$\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34$ Solving for H:

$H=\sqrt{52}$

We can now, calculate the angle between L1 and the 2m segment:

$\alpha = atan(\frac{H/2}{2})=60.98°$

If we make a sum of forces in the midpoint of the rope we get:

$-2*T*cos(\alpha ) + F = 0$ where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

$T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N$

7 0

3 years ago