All categories

3 years ago

पुस्तक विक्रेता ने वीपीपी द्वारा आपको पुस्तक भेजी है?? धन्यवाद देते हुए पत्र लिखिए।

Physics

1 answer:

Xelga [282]3 years ago

3 0

Answer:

y e s :))))))))))))))))))))))))))))))))

Explanation:

y - dirdf leter of di alfabet

You might be interested in

Help please me please

Colt1911 [192]

B

add them all by direction
13 East
10 West
subtract difference
3 E

8 0

3 years ago

How will decreasing the amplitude of the sound waves from a television affect its intensity?

Colt1911 [192]

A. The sound will decrease in volume

6 0

3 years ago

Find the magnitude and direction of an electric field that exerts a 4.80×10−17N westward force on an electron. (b) What magnitud

jok3333 [9.3K]

Explanation:

(a) E = F/q

E = 4.8×10^-17/1.6×10^-19

E = 300 N/C

(b) same magnitude of electric field is exerted on proton

4 0

3 years ago

A person swings a 0.546-kg tether ball tied to a 4.56-m rope in an approximately horizontal circle. If the maximum tension the r

Murrr4er [49]

Answer:

2.1 rad/s

Explanation:

Given that,

Mass of a tether ball, m = 0.546 kg

Length of a rope, l = 4.56 m

The maximum tension the rope can withstand before breaking is 11.0 N

We need to find the maximum angular speed of the ball. Let v is the linear velocity. The maximum tension is balanced by the centripetal force acting on it. It can be given by :

$F=\dfrac{mv^2}{r}\\\\v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{11\times 4.56}{0.546}} \\\\=9.584\ m/s$

Let $\omega$ is the angular speed of the ball. The relation between the angular speed and angular velocity is given by :

$v=r\omega\\\\\omega=\dfrac{v}{r}\\\\=\dfrac{9.584}{4.56}\\\\=2.1\ rad/s$

So, the maximum angular speed of the ball is 2.1 rad/s.

4 0

3 years ago

A block of a plastic material floats in water with 42.9% of its volume under water. What is the density of the block in kg/m3?

adell [148]

To solve this problem we will apply the principle of buoyancy of Archimedes and the relationship given between density, mass and volume.

By balancing forces, the force of the weight must be counteracted by the buoyancy force, therefore

$\sum F = 0$

$F_b -W = 0$

$F_b = W$

$F_b = mg$

Here,

m = mass

g =Gravitational energy

The buoyancy force corresponds to that exerted by water, while the mass given there is that of the object, therefore

$\rho_w V_{displaced} g = mg$

Remember the expression for which you can determine the relationship between mass, volume and density, in which

$\rho = \frac{m}{V} \rightarrow m = V\rho$

In this case the density would be that of the object, replacing

$\rho_w V_{displaced} g = V\rho g$

Since the displaced volume of water is 0.429 we will have to

$\rho_w (0.429V) = V \rho$

$0.429\rho_w= \rho$

The density of water under normal conditions is $1000kg / m ^ 3$ , so

$0.429(1000) = \rho$

$\rho = 429kg/m^3$

The density of the object is $429kg / m ^ 3$

7 0

3 years ago