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1 year ago

Hydraulic press is called an instrument for multiplication of force. Why?

Physics

1 answer:

Lisa [10]1 year ago

4 0

Answer:

Hydraulic press is called an instrument for multiplication of force. Why? Because it uses Pascal's idea and principle: F=p*S. If we apply small force to small piston you will generate a pressure. According to Pascal's law pressure is the same everywhere in closed system so the same pressure will act on large piston on the other side too.

Explanation:

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A parallel-plate capacitor with a 4.9 mm plate separation is charged to 57 V . Part A With what kinetic energy, in eV, must a pr

Dovator [93]

Answer:

57 eV

Explanation:

$d$ = separation between the plates = 4.9 mm = 0.0049 m

$\Delta V$ = Potential difference between the plates = 57 Volts

$q$ = magnitude of charge on proton = 1 e

$K$ = Kinetic energy of the proton

Using conservation of energy

Kinetic energy lost = Electric potential energy gained

$K = q \Delta V\\K = (1 e) (57 )\\K = 57 eV$

4 0

2 years ago

A charge of 4 nc is placed uniformly on a square sheet of nonconducting material of side 17 cm in the yz plane. (a) what is the

Ratling [72]

The charge density of the sheet is 1.384×10⁻⁷C/m².

Charge density is defined as the charge per unit area.

The sheet is a square of length l=17 cm.

Calculate the area A of the sheet .

$A=l^2=(17 cm)^2= (17*10^-^2m)^2=0.0289 m^2$

The charge Q on the sheet is

$Q=4nC=4*10^-^9C$

The charge density σ is given by,

$\sigma=\frac{Q}{A}$

Substitute 4×10⁻⁹C for Q and 0.0289 m² for A.

$\sigma=\frac{Q}{A}\\ =\frac{4*10^-^9C}{0.0289 m^2} \\ =1.389*10^-^7C/m^2$

Thus, the charge density of the sheet is <u>1.384×10⁻⁷C/m².</u>

8 0

2 years ago

ery large accelerations can injure the body, especially if they last for a considerable length of time. One model used to gauge

IgorC [24]

Answer:

2.98 second

Explanation:

The severity index is defined by :

$S=a^{5/2}t$

a is dimensionless constant that equals the number of multiples of g

Conditions are given as :

Initial velocity, u = 0

Acceleration, a = 34 m/s²

Final velocity, v = 16.4 km/h = 4.56 m/s

We can find t from the above data as follows :

$t=\dfrac{v-u}{a}\\\\t=\dfrac{4.56-0}{34}\\\\t=0.134\ s$

As a is the acceleration that is multiple of g.

So,

$a=\dfrac{34}{9.8}=3.46$

So,

Severity index,

$S=a^{5/2}t\\\\S=(3.46)^{5/2}\times 0.134\\\\S=2.98\ s$

Hence, the severity index for the collision is 2.98 seconds.

6 0

2 years ago

A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air

Alex Ar [27]

Answer:

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball, $v_{o} = 20.0 m/s$

Angle made by the ramp, $\theta = 22.0^{\circ}$

Distance traveled by the ball on the ramp, d = 5.00 m

Now,

(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:

$v^{2} = v_{o}^{2} - 2gH$

where

H = $dsin22^{\circ} = 5sin22^{\circ}$

g = $9.8 m/s^{2}$

$v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ}$

$v = \sqrt{400 - 19.6\times 5sin22^{\circ}}$ = 19.06 m/s

Now, maximum height attained is given by:

$h = \frac{(vsin\theta)^{2}}{2g}$

$h = \frac{(19sin(22^{\circ}))^{2}}{2\times 9.8} = 2.60 m$

Height from the ground = $5sin22^{circ} + 2.86 = 1.87 + 2.60 = 4.47m$

(b) now, considering the coefficient of friction bhetween ramp and the ball, $\mu = 0.150$ :

velocity can be given by the eqn-3 of motion:

$v^{2} = v_{o}^{2} - 2gH - \mu gd$

$v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ} - 0.150\times 9.8\times 5$

$v = \sqrt{400 - 19.6\times 5sin22^{\circ} - 0.150\times 9.8\times 5}$ = 18.7 m/s

Now, maximum height attained is given by:

$h = \frac{(vsin\theta)^{2}}{2g}$

$h = \frac{(18.7sin(22^{\circ}))^{2}}{2\times 9.8} = 2.50 m$

Height from the ground = $5sin22^{circ} + 2.86 = 1.87 + 2.50 = 4.37 m$

6 0

1 year ago

In a scientific test conducted in Arizona, a special cannon called HARP (High Altitude Research Project) shot a projectile strai

Alexus [3.1K]

Answer:

It took the projectile 120 s to reach the maximum height.

Explanation:

Given;

maximum height of the projectile, s = 180 km = 180,000 m

initial speed of the projectile, u = 3 km/s = 3000 m/s

final velocity at maximum height, v = 0

Apply the following kinematic equation for average velocity of the projectile;

$s = (\frac{v+u}{2} )t\\\\(v+u)t = 2s\\\\t = \frac{2s}{v+u} \\\\t = \frac{2*180,000}{0+3,000}\\\\ t = 120 \ s$

Therefore, it took the projectile 120 s to reach the maximum height.

5 0

2 years ago