All categories

3 years ago

A box is 30cm wide , 50cm long and 20cm high Calculate :

Physics

1 answer:

stellarik [79]3 years ago

7 0

Answer:

A) 1500cm ^2

B) 30000 CM ^3

You might be interested in

Match the following. Column A 1. Torque 2. Centre of gravity 3. Plumb line Column B A. Line of centre of gravity B. Maximum cons

gregori [183]

Answer:

1. Torque → F. Study of forces

2. C.O.G → D. Point of action of weight.

3. Plumb line → A. Line of C.O.G

8 0

3 years ago

A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium whe

ArbitrLikvidat [17]

Answer: Here is the complete question:

A small 12.00g plastic ball is suspended by a string in a uniform, horizontal electric field with a magnitude of 103 N/C. If the ball is in equilibrium when the string makes a 30 angle with the vertical, what is the net charge on the ball?

Answer: The charge on the ball is 5.71 × 10^-4 C

Explanation:

Please see the attachments below

5 0

3 years ago

A skydiver of mass 80kg jumps from a slow moving aircraft and reach a terminal speed of 50 m/s. What's her acceleration when her

miskamm [114]

Answer:

a = g = 9.81[m/s^2]

Explanation:

This problem can be solve using the second law of Newton.

We know that the forces acting over the skydiver are only his weight, and it is equal to the product of the mass by the acceleration.

m*g = m*a

where:

g = gravity = 9.81[m/s^2]

a = acceleration [m/s^2]

Note: If the skydiver will be under air resistance forces his acceleration will be different.

7 0

3 years ago

A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn

Rus_ich [418]

Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J

Explanation:

We have given the player turntable initially rotating at speed of $33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec$

Now speed is reduced by 75 %

So final speed $\frac{3.49\times 75}{100}=2.6175rad/sec$

Time t = 5.5 sec

From first equation of motion we know that '

$\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2$

(a) Now final velocity $\omega =0rad/sec$

So time t to come in rest $t=\frac{0-3.49}{-0.158}=22sec$

(b) The work done in coming rest is given by

$\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J$

4 0

2 years ago

A 200g block on a 50cm long string swings in a circle, it's frictionless and 75rpm. What is its speed and tension on string

krek1111 [17]

Angular velocity = (75x2pie)/60
      =2.5pie ras^-1
linear velocity(or speed) at end of string, v = radius x angular velocity
       v= 0.5 x 2.5pie
   v=3.93 ms^-1

tension of string (I beleve is centeral force aplied by string), F= (mv^2)/r
      F= (0.2 x 3.93^2)/0.5
      F=6.18 N
(sorry if wrong)

3 0

2 years ago

Read 2 more answers