Does an object have energy when it is at rest?

A rectangular tank is filled with water to a depth of 4m. Calculate the pressure at the bottom of the tank. PLEASEEE HELP!!

Paul [167]

The pressure in a liquid at a given depth is called the hydrostatic pressure. This can be calculated using the hydrostatic equation:

P = ρgh

P = (1000 kg/m³)(9.8 m/s²)(4 m)

4 0

3 years ago

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. An object has a position given by ~r(t) = [3.0 m − (4.00 m/s)t]ˆı + [6.0 m − (8.00 m/s2 )t 2 ]ˆ , where all quantities are in

kupik [55]

Answer:

(c) 16 m/s²

Explanation:

The position is $r(t) = [3.0 \text{ m} - (4.00 \text{ m/s})t]\hat{i} + [6.0 \text{m} - (8.00 \text{ m/s}^2 )t^2 ]\hat{j}$ .

The velocity is the first time-derivative of r(t).

$v(t) = \dfrac{d}{dt}r(t) = -4.00\,\hat{i} -16t\,\hat{j}$

The acceleration is the first time-derivative of the velocity.

$a(t) = \dfrac{d}{dt} v(t) = -16\hat{j}$

Since a(t) does not have the variable t, it is constant. Hence, at any time,

$a = -16\hat{j}$

Its magnitude is 16 m/s².

4 0

3 years ago

An unknown charged particle passes without deflection throughcrossed electric and magnetic fields of strengths 187,500 V/m and0.

UNO [17]

Explanation:

The given data is as follows.

Electric field strength (E) = 187,500 V/m

Magnetic field strength (B) = 0.125 T

Diameter (d) = 25.05 cm = 0.2505 m (as 1 m = 100 cm)

Radius (r) = $\frac{d}{2}$

= $\frac{0.2505}{2}$

= 0.12525 m

Formula to calculate the magnetic force ( $F_{M}$ ) is as follows.

$F_{M} = Bqv$ ............ (1)

Electrical force is calculated as follows.

$F_{E} = qE$ ............ (2)

On both electric and magnetic fields the velocity is perpendicular.

$F_{M} - F_{E}$ = 0

or, $F_{M} = F_{E}$

Hence, from equations (1) and (2)

Bqv = qE

or, v = $\frac{E}{B}$ ............. (3)

= $\frac{187500 V/m}{0.125 T}$

= 1,500,000 m/s

As the particle is moving in a semi-circular trajectory and motion of charged particle is given by the electric field as follows.

$F_{c} = \frac{mv^{2}}{r}$ ........... (4)

where, $F_{c}$ = centripetal force

$F_{M} = F_{c}$

Using equation (1) and (4) as follows.

$F_{M} = F_{c}$

Bqv = $\frac{mv^{2}}{r}$

$\frac{q}{m} = \frac{v}{Br}$

= $\frac{15 \times 10^{5}}{0.125 \times 0.12525}$

= $958.08 \times 10^{5} C/kg$

Thus, we can conclude that charge-to-mass ratio of the given particle is $958.08 \times 10^{5} C/kg$ .

8 0

4 years ago

Could anyone help me with this question?

Nana76 [90]

The last 2. They both are correct. I hope this helps.

3 0

3 years ago

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"On a movie set, an alien spacecraft is to be lifted to a height of 32.0 m for use in a scene. The 260.0-kg spacecraft is attach

Lisa [10]

Answer:

The time interval required to lift the spacecraft to this specified height is 123.94 seconds

Explanation:

Height through which the spacecraft is to be lifted = 32.0 m

Mass of the spacecraft = 260.0 kg

Four crew member each pull with a power of 135 W

18.0% of the mechanical energy is lost to friction.

work done in this situation is proportional to the mechanical energy used to move the spacecraft up

work done = (weight of spacecraft) x (the height through which it is lifted)

but the weight of spacecraft = mg

where m is the mass,

and g is acceleration due to gravity 9.81 m/s

weight of spacecraft = 260 x 9.81 = 2550.6 N

work done on the space craft = weight x height

==> work = 2550.6 x 32 = 81619.2 J

this is equal to the mechanical energy delivered to the system

18.0% of this mechanical energy delivered to the pulley is lost to friction.

this means that

0.18 x 81619.2 = 14691.456 J is lost to friction.

Total useful mechanical energy = 81619.2 J - 14691.456 J = 66927.74 J

Total power delivered by the crew to do this work = 135 x 4 = 540 W

But we know tat power is the rate at which work is done i.e

$P = \frac{w}{t}$

where p is the power

where w is the useful work done

t is the time taken to do this work

imputing values, we'll have

540 = 66927.74/t

t = 66927.74/540

time taken t = 123.94 seconds

8 0

4 years ago