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2 years ago

A rock is lying on a rock ledge that is 3 m high. The rock as 120 J of potential energy. What is the mass of the rock?

Physics

1 answer:

Sonja [21]2 years ago

6 0

$\\ \sf\Rrightarrow PE=mgh$

$\\ \sf\Rrightarrow m(10)(3)=120$

$\\ \sf\Rrightarrow 30m=120$

$\\ \sf\Rrightarrow m=4kg$

Mass=4kg

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A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball

Akimi4 [234]

Answer:

Approximately $122.625\; {\rm m}$ (assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$ , the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let $v_{0}$ denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly $(-v_{0})$ . The velocity of the ball would be changed from $v$ to $(-v_{0})\!$ (such that $\Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})$ ) within $t = 10\; {\rm s}$ .

Also because the drag on the ball is negligible, the acceleration of the ball would be $a = -g = -9.81\; {\rm m\cdot s^{-2}}$ . Thus:

$\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}$ .

Since $\Delta v = (-2\, v_{0})$ :

$-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}$ .

$\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}$ .

The ball reaches maximum height when its velocity is $v_{1} = 0\; {\rm m\cdot s^{-1}}$ . Apply the SUVAT equation $x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a)$ to find the displacement $x$ between the original position (ground level, where $v_{0} = 49.05\; {\rm m\cdot s^{-1}}$ ) and the max-height position of the ball (where $v_{1} = 0\; {\rm m\cdot s^{-1}}$ .)

$\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}$ .

7 0

2 years ago

The wavelength of red light is 7 x 10^-7 meter. Express this value in nanometers

3241004551 [841]

The wavelength of the red light in "nanometer" is 7× $10^{2}$

Wavelength is given as : 7× $10^{-7}$ meter

1 nanometer = ( $10^{-9}$ meter)

Let X= value of the wavelength in nanometer.

1 nanometer = $10^{-9}$ meter

X nanometer = 7× $10^{-7}$ meter

<em>If we Cross multiply</em>

X nanometer = ( $\frac{ 7* 10^{-7} }{ 10^{-9} }$ )

X= 7× $10^{2}$ nanometer

Therefore, the wavelength in "nanometer" is 7× $10^{2}$

Learn more at :brainly.com/question/12924624?referrer=searchResults

6 0

3 years ago

If one 9V battery is used in a circuit with a total resistance of 39Ω, what is the current in the circuit?

Ket [755]

Answer:

Using V= IR

I= 0.2307 Ampere

6 0

2 years ago

What is the area of the bottom of a tank 30.0 cm long and 15.0 cm wide?

Igoryamba

Answer:

perimeter of the bottom of the tank is 450 cm

Explanation:

If you want to get area you need the Length, width,and height.

5 0

3 years ago

Read 2 more answers

The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from

Allushta [10]

Answer:

y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

x = v₀ₓ t

y = $v_{oy}$ t - ½ g t²

let's substitute

45 = v₀ cos θ t

10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

v_{y} = v_{oy} - gt

0 = v₀ sin θ - gt

0 = v₀ sin θ - 9.8 t

Let's write our system of equations

45 = v₀ cos θ t

10 = v₀ sin θ t - ½ 9.8 t²

0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

v₀ sin θ = 9.8 t

we substitute

10 = (9.8 t) t - ½ 9.8 t2

10 = ½ 9.8 t2

10 = 4.9 t2

t = √ (10 / 4.9)

t = 1,429 s

Now let's use the first equation and the last one

45 = v₀ cos θ t

0 = v₀ sin θ - 9.8 t

9.8 t = v₀ sin θ

45 / t = v₀ cos θ

we divide

9.8t / (45 / t) = tan θ

tan θ = 9.8 t² / 45

θ = tan⁻¹ ( 9.8 t² / 45 )

θ = tan⁻¹ (0.4447)

θ = 24º

Now we can calculate the maximum height

v_y² = $v_{oy}^2$ - 2 g y

vy = 0

y = v_{oy}^2 / 2g

y = (20 sin 24)²/2 9.8

y = 3,376 m

the other angle that gives the same result is

θ‘= 90 - θ

θ' = 90 -24

θ'= 66'

for this angle the maximum height is

y = v_{oy}^2 / 2g

y = (20 sin 66)²/2 9.8

y = 17 m

thisis the correct

6 0

3 years ago