Does uniform acceleration means increasing acceleration?

A 2,000 kg car is parked at the top of a 30 m high hill. What is its potential

Lunna [17]

So the right answer is of option D .

Look at the attached picture

Hope it will help you ...

Good luck on your assignment

~pragya

4 0

3 years ago

The water behind Grand Coulee Dam is 1000 m wide and 200 m deep. Find the hydrostatic force on the back of the dam. (Hint: the t

Vika [28.1K]

Answer:

The hydro static force on the back of the dam is $1.96\times10^{11}\ N$

Explanation:

Given that,

Width b= 1000 m

Depth d= 200 m

We need to calculate the average pressure

Using formula of average pressure

$P_{avg}=\rho\times g\times d_{avg}$

Put the value into the formula

$P_{avg}=1000\times9.8\times100$

$P_{avg}=980000\ Pa$

We need to calculate the hydro static force on the back of the dam

Using formula of force

$F = P_{avg}\times A$

Put the value into the formula

$F = 980000\times1000\times200$

$F=1.96\times10^{11}\ N$

Hence, The hydro static force on the back of the dam is $1.96\times10^{11}\ N$

7 0

2 years ago

Plz help I will give brainly to the correct answer

Tresset [83]

Answer:

D. x-rays

Explanation:

Lower frequency: Radio waves, microwaves and infrared have lower frequency than visible light. Shorter wavelength: Ultraviolet, x-rays and gamma rays have a shorter wavelength than visible light.

4 0

2 years ago

The current theory of the structure of the Earth, called plate tectonics, tells us that the continents are in constant motion.

lesya [120]

Answer:

(a) m = 1.6 x 10²¹ kg

(b) K.E = 2.536 x 10¹¹ J

(c) v = 7.12 x 10⁵ m/s

Explanation:

(a)

First we find the volume of the continent:

V = L*W*H

where,

V = Volume of Slab = ?

L = Length of Slab = 4450 km = 4.45 x 10⁶ m

W = Width of Slab = 4450 km = 4.45 x 10⁶ m

H = Height of Slab = 31 km = 3.1 x 10⁴ m

Therefore,

V = (4.45 x 10⁶ m)(4.45 x 10⁶ m)(3.1 x 10⁴ m)

V = 6.138 x 10¹⁷ m³

Now, we find the mass:

m = density*V

m = (2620 kg/m³)(6.138 x 10¹⁷ m³)

m = 1.6 x 10²¹ kg

(b)

The kinetic energy will be:

K.E = (1/2)mv²

where,

v = speed = (1 cm/year)(0.01 m/1 cm)(1 year/365 days)(1 day/24 h)(1 h/3600 s)

v = 3.17 x 10⁻¹⁰ m/s

Therefore,

K.E = (1/2)(1.6 x 10²¹ kg)(3.17 x 10⁻¹⁰ m/s)²

K.E = 2.536 x 10¹¹ J

(c)

For the same kinetic energy but mass = 77 kg:

K.E = (1/2)mv²

2.536 x 10¹¹ J = (1/2)(77 kg)v²

v = √(2)(2.536 x 10¹¹ J)

v = 7.12 x 10⁵ m/s

7 0

2 years ago

A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−

Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ .

Explanation:

Given that,

The actual wavelength of the hydrogen atom, $\lambda_a=5.13\times 10^{-7}\ m$

A hydrogen atom in a galaxy moving with a speed of, $v=6.65\times 10^6\ m/s$

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

$v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}$

$\lambda_o$ is the observed wavelength

$\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m$

So, the observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ . Hence, this is the required solution.

8 0

2 years ago