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Liono4ka [1.6K]

3 years ago

14

A spring stretches by 0.014 m when a 3.7-kg object is suspended from its end. How much mass should be attached to this spring so

that its frequency of vibration is f

1 answer:

lidiya [134]3 years ago

6 0

Explanation:

Below is an attachment containing the solution.

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What is the gravitational force between two students, John and Mike, if John has a mass of 81.0 kg, Mike has a mass of 93.0 kg,

Marianna [84]

Answer:

1.31×10¯⁶ N

Explanation:

From the question given above, the following data were obtained:

Mass of John (M₁) = 81 Kg

Mass of Mike (M₂) = 93 Kg

Distance apart (r) = 0.620 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force (F) =?

The gravitational force between the two students, John and Mike, can be obtained as follow:

F = GM₁M₂ / r²

F = 6.67×10¯¹¹ × 81 × 93 / 0.62²

F = 6.67×10¯¹¹ × 7533 / 0.3844

F = 1.31×10¯⁶ N

Therefore, the gravitational force between the two students, John and Mike, is 1.31×10¯⁶ N

8 0

2 years ago

URGENT!!!!!!!<br><br> PLEASE HELP WITH THIS PHYSICS PROBLEM

Levart [38]

Explanation:

Let

$x_1$ = distance traveled while accelerating

$x_2$ = distance traveled while decelerating

The distance traveled while accelerating is given by

$x_1 = v_0t + \frac{1}{2}at^2 = \frac{1}{2}at^2$

$\:\:\:\:\:= \frac{1}{2}(2.5\:\text{m/s}^2)(30\:\text{s})^2$

$\:\:\:\:\:= 1125\:\text{m}$

We need the velocity of the rocket after 30 seconds and we can calculate it as follows:

$v = at = (2.5\:\text{m/s}^2)(30\:\text{s}) = 75\:\text{m/s}$

This will be the initial velocity when start calculating for the distance it traveled while decelerating.

$v^2 = v_0^2 + 2ax_2$

$0 = (75\:\text{m/s})^2 + 2(-0.65\:\text{m/s}^2)x_2$

Solving for $x_2,$ we get

$x_2 = \dfrac{(75\:\text{m/s})^2}{2(0.65\:\text{m/s}^2)}$

$\:\:\:\:\:= 4327\:\text{m}$

Therefore, the total distance x is

$x = x_1 + x_2 = 1125\:\text{m} + 4327\:\text{m}$

$\:\:\:\:= 5452\:\text{m}$

3 0

3 years ago

A car going initially with a velocity 13.5 m/s accelerates at a rate of 1.9 m/s for 6.2 s. It then accelerates at a rate of-1.2

yanalaym [24]

Answer:

a) Maximum speed = 25.28 m/s

b) Total time = 27.27 s

c) Total distance traveled = 402.43 m

Explanation:

a) Maximum speed is obtained after the end of acceleration

v = u + at

v = 13.5 + 1.9 x 6.2 = 25.28 m/s

Maximum speed = 25.28 m/s

b) We have maximum speed = 25.28 m/s, then it decelerates 1.2 m/s² until it stops.

v = u + at

0 = 25.28 - 1.2 t

t = 21.07 s

Total time = 6.2 + 21.07 = 27.27 s

c) Distance traveled for the first 6.2 s

s = ut + 0.5 at²

s = 13.5 x 6.2 + 0.5 x 1.9 x 6.2² = 120.22 m

Distance traveled for the second 21.07 s

s = ut + 0.5 at²

s = 25.28 x 21.07 - 0.5 x 1.2 x 21.07² = 282.21 m

Total distance traveled = 120.22 + 282.21 = 402.43 m

5 0

2 years ago

Read 2 more answers

To suck lemonade of density 1040 kg/m3 up a straw to a maximum height of 4.94 cm, what minimum gauge pressure (in atmospheres) m

Lady bird [3.3K]

Answer:

The minimum gauge pressure is 0.4969 atm.

Explanation:

Given that,

Density = 1040 kg/m³

Height = 4.94 cm

We need to calculate the pressure

Using formula of pressure

$P_{g}=\rho g h$

Where, $\rho$ =density

h = height

Put the value into the formula

$P_{g}=1040\times9.8\times4.94$

$P_{g}=50348.48\ Pa$

Pressure in atmospheres

$1\ atm =101.3\ kPa$

$P_{g}=\dfrac{50348.48}{101325}$

$P_{g}=0.4969\ atm$

Hence, The minimum gauge pressure is 0.4969 atm.

7 0

3 years ago

Read 2 more answers

What is the maximum value of the magnetic field at a distance of 2.5 m from a light bulb that radiates 100 W of single-frequency

Anvisha [2.4K]

Answer:

$1.04\times 10^{-7}$ T

Explanation:

$IP$ = Power of the bulb = 100 W

$r$ = distance from the bulb = 2.5 m

$I$ = Intensity of light at the location

Intensity of the light at the location is given as

$I = \frac{P}{4\pi r^{2}}$

$I = \frac{100}{4(3.14) (2.5)^{2}}$

$I$ = 1.28 W/m²

$B_{o}$ = maximum magnetic field

Intensity is given as

$I = \frac{B_{o}^{2}c}{2\mu _{o}}$

$1.28 = \frac{B_{o}^{2}(3\times 10^{8})}{2(12.56\times 10^{-7})}$

$B_{o} = 1.04\times 10^{-7}$ T

7 0

3 years ago