All categories

3 years ago

What are the conditions for a body to be at equilibrium

2 answers:

6 0

Conditions for equilibrium require that the sum of all external forces acting on the body is zero (first condition of equilibrium), and the sum of all external torques from external forces is zero (second condition of equilibrium). These two conditions must be simultaneously satisfied in equilibrium.

uranmaximum [27]3 years ago

4 0

Answer:

Explanation:

You might be interested in

At high speeds, a particular automobile is capable of an acceleration of about 0.540 m/s^2. At this rate, how long (in seconds)

Mama L [17]

Answer:

t = 6.68 seconds

Explanation:

The acceleration of the automobile, $a=0.54\ m/s^2$

Initial speed of the automobile, u = 91 km/hr = 25.27 m/s

Final speed of the automobile, v = 104 km/hr = 28.88 m/s

Let t is the time taken to accelerate from u to v. It can be calculated as the following formula as :

$t=\dfrac{v-u}{a}$

$t=\dfrac{28.88-25.27}{0.54}$

t = 6.68 seconds

So, the time taken by the automobile to accelerate from u to v is 6.68 seconds. Hence, this is the required solution.

5 0

3 years ago

A daring stunt woman sitting on a tree limb wishes to drop vertically onto a horse gallop ing under the tree. The constant speed

Shkiper50 [21]

Answer:

0.67 seconds

8.576 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.23=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2.23\times 2}{9.8}}\\\Rightarrow t=0.67\ s$

Time taken by the stunt woman to drop to the saddle is 0.67 seconds which is the time she will stay in the air.

Speed of the horse = 12.8 m/s

Distance = Speed × Time

⇒Distance = 12.8×0.67

⇒Distance = 8.576 m

Hence, the distance between the horse and stunt woman should be 8.576 m when she jumps.

3 0

3 years ago

Which of these statements is true?

9966 [12]

C.) Meiosis involves two cycles of cell division

Hope this helps!

6 0

4 years ago

Read 2 more answers

What do you think would happen to the force of attraction of two interacting charges if their distance apart is halved?

sweet [91]

Answer:

The new force becomes 4 times the initial force.

Explanation:

The force of attraction or repulsion is given by the relation as follows :

$F=k\dfrac{q_1q_2}{d^2}$

Where

d is the distance between the interacting charges

F is inversely proportional to the distance between charges.

If the distance is halved, d'=(d/2), new force is given by :

$F'=k\dfrac{q_1q_2}{d'^2}\\\\=k\dfrac{q_1q_2}{(\dfrac{d}{2})^2}\\\\=k\dfrac{q_1q_2}{\dfrac{d^2}{4}}\\\\=4\times \dfrac{kq_1q_2}{d^2}\\\\F'=4F$

So, the new force becomes 4 times the initial force.

4 0

3 years ago

Need answers quick !! will give brainliest !!

stiv31 [10]

Answer:

use the bowl of water because Earth's magnetic field is relatively weak. Allowing it to float freely on the water, allows the magnetized needle to freely react to Earth's magnetic field, causing it to align North to South. If you watched closely, the same end of the needle should always point to the North

Explanation:

6 0

3 years ago