Help please !! physics kinematic equations

A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi

Dmitriy789 [7]

Answer:

a

$A = 0.081 \ m$

b

The value is $u = 0.2569 \ m/s$

Explanation:

From the question we are told that

The mass is $m = 0.750 \ kg$

The spring constant is $k = 17.5 \ N/m$

The instantaneous speed is $v = 39.0 \ cm/s= 0.39 \ m/s$

The position consider is x = 0.750A meters from equilibrium point

Generally from the law of energy conservation we have that

The kinetic energy induced by the hammer = The energy stored in the spring

So

$\frac{1}{2} * m * v^2 = \frac{1}{2} * k * A^2$

Here a is the amplitude of the subsequent oscillations

=> $A = \sqrt{\frac{m * v^ 2 }{ k} }$

=> $A = \sqrt{\frac{0.750 * 0.39 ^ 2 }{17.5} }$

=> $A = 0.081 \ m$

Generally from the law of energy conservation we have that

The kinetic energy by the hammer = The energy stored in the spring at the point considered + The kinetic energy at the considered point

$\frac{1}{2} * m * v^2 = \frac{1}{2} * k x^2 + \frac{1}{2} * m * u^2$

=> $\frac{1}{2} * 0.750 * 0.39^2 = \frac{1}{2} * 17.5* 0.750(0.081 )^2 + \frac{1}{2} * 0.750 * u^2$

=> $u = 0.2569 \ m/s$

3 0

3 years ago

1. An electron (Q=16x10^-20 C, m=1x10^-30 kg) moving at half a megameter per second up the page enters a region with a uniform m

jek_recluse [69]

Explanation:

It is given that,

Charge on electron, $q=16\times 10^{-20}\ C$

Mass of the electron, $m=9.1\times 10^{-31}\ kg$

Speed of the electron, $v=0.5\ Mm/s=0.5\times 10^6\ m/s$

Magnetic field, B = 1 T (directed out of the page)

Let F is the magnetic force acting on the electron. It is given by :

$F=qvB\ sin\theta$

Here, $\theta=90^{\circ}$

$F=qvB$

$F=16\times 10^{-20}\ C\times 0.5\times 10^6\ m/s\times 1\ T$

$F=8\times 10^{-14}\ N$

Using the right hand rule, the direction of magnetic force is upward to the plane of the paper. Also, the electron will follow the circular path. It is given by :

$r=\dfrac{mv}{qB}$

$r=\dfrac{9.1\times 10^{-31}\ kg\times 0.5\times 10^6\ m/s}{16\times 10^{-20}\ C\times 1\ T}$

$r=2.84\times 10^{-6}\ m$

Hence, this is the required solution.

7 0

3 years ago

It takes me 12s to lift a box upward with 20 N of force to a height of 45m. What was my power output ?

Finger [1]

Answer:

75Watts

Explanation:

Given parameters:

Time = 12s

Force applied = 20N

Height = 45m

Unknown:

Power output = ?

Solution:

Power is defined as the rate at which work is done.

It is mathematically expressed as;

Power = $\frac{work done}{time}$

Work done = force x distance = 20 x 45 = 900J

Power = $\frac{900}{12}$ = 75Watts

6 0

3 years ago

A standard light bulb emits light rays in _________ directions.

Ainat [17]

Omni-directional. It means all directional.

4 0

3 years ago

On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3 , length 92.6 cm and diameter 2.95 cm fr

slavikrds [6]

Answer:

48.4293354946 N

Yes

Explanation:

d = Diameter of rod = 2.95 cm

h = Length of rod = 92.6 cm

$\rho$ = Density of rod = 7800 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Volume of rod

$V=\dfrac{1}{4}\pi d^2h\\\Rightarrow V=\dfrac{1}{4}\times \pi\times (2.95\times 10^{-2})^2\times 92.6\times 10^{-2}$

Mass is given by

$m=\rho V\\\Rightarrow m=7800\times \dfrac{1}{4}\times \pi\times (2.95\times 10^{-2})^2\times 92.6\times 10^{-2}\\\Rightarrow m=4.93673144695\ kg$

Weight is given by

$W=mg\\\Rightarrow W=4.93673144695\times 9.81\\\Rightarrow W=48.4293354946\ N$

The weight of the rod is 48.4293354946 N

The mass of the rod is 4.93673144695 kg which is light. So, I will be able to carry the rod without a cart.

7 0

3 years ago