All categories

2 years ago

If a 1,300 kg car with no people inside is on the edge of a cliff 1,500 m above the ground, what is its potential energy?

Physics

1 answer:

Ghella [55]2 years ago

3 0

<u>Given that</u>

mass (m) = 1300 Kg ,

height (h) = 1500 m

Determine the potential energy ?

P.E = m × g × h

= 1300 × 9.81 × 1500

= 19129500 Joules

= 19129.5 KJ

You might be interested in

A spherical balloon is 40 ft in diameter and surrounded by air at 60°F and 29.92 in Hg abs.(a) If the balloon is filled with hyd

masha68 [24]

It's obviously b because the more helium it has, the lighter it is...

3 0

2 years ago

Read 2 more answers

A bowling ball of mass m=1.7kg is launched from a spring compressed by a distance d=0.31m at an angle of theta=37 measured from

vodomira [7]

Answer:

k = 1 700.7 N/m

v0 = 9.8 m/s^2

Explanation:

Hello!

We can answer this question using conservation of energy.

The potential energy of the spring (PS) will transform to kinetic energy (KE) of the ball, and eventually, when the velocity of the ball is zero, all that energy will be potential gravitational (PG) energy.

When the kinetic energy of the ball is zero, that is, when it has reached its maximum heigh, all the potential energy of the spring will be equal to the potential energy of the gravitational field.

PS = (1/2) k x^2 <em>where x is the compresion or elongation of the spring</em>

PG = mgh

Since energy must be conserved and we are neglecting any energy loss:

PS = PG

Solving for k

k = (2mgh)/(x^2) = ( 2 * 1.7 * 9.81 * 4.9 Nm)/(0.31^2 m^2)

k = 1 700.7 N/m

Since the potential energy of the spring transfors to kinetic energy of the ball we have that:

PS = KE

that is:

(1/2) k x^2 = (1/2) m v0^2

Solving for v0

v0 = x √(k/m) = (0.31 m ) √( 1 700.7 N/m / 1.7kg)

v0 = 9.8 m/s^2

8 0

2 years ago

An electron moves in a region where the magnetic field is uniform and has a magnitude of 80 μT. The electron follows a helical p

sladkih [1.3K]

Answer:

3.4 x 10⁴ m/s

Explanation:

Consider the circular motion of the electron

B = magnetic field = 80 x 10⁻⁶ T

m = mass of electron = 9.1 x 10⁻³¹ kg

v = radial speed

r = radius of circular path = 2 mm = 0.002 m

q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

For the circular motion of electron

qBr = mv

(1.6 x 10⁻¹⁹) (80 x 10⁻⁶) (0.002) = (9.1 x 10⁻³¹) v

v = 2.8 x 10⁴ m/s

Consider the linear motion of the electron :

v' = linear speed

x = horizontal distance traveled = 9 mm = 0.009 m

t = time taken = $\frac{2\pi m}{qB}$ = $\frac{2\pi (9.1\times 10^{-31})}{(1.6\times 10^{^{-19}})(80\times 10^{-6})}$ = 4.5 x 10⁻⁷ sec

using the equation

x = v' t

0.009 = v' (4.5 x 10⁻⁷)

v' = 20000 m/s

v' = 2 x 10⁴ m/s

Speed is given as

V = sqrt(v² + v'²)

V = sqrt((2.8 x 10⁴)² + (2 x 10⁴)²)

v = 3.4 x 10⁴ m/s

6 0

2 years ago

What is NOT one of the three primary resources that families have to reach financial goals?

Elodia [21]

What is NOT one of the three primary resources that families have to reach financial goals? It is c) education

8 0

2 years ago

A person fishing from a pier observes that 6 wave crests pass by in 8.0 s and estimates the distance between two successive cres

Talja [164]

Answer:

The speed of the wave is 3 m/s.

Explanation:

Given that,

Number of waves = 6

Time = 8.0 s

Successive crests $\lambda = 4.8\ m$

We need to calculate the frequency

Using formula of frequency

$f=\dfrac{Number\ of \ cycles}{t}$

Put the value into the formula

$f=\dfrac{6-1}{8.0}$

$f=0.625\ cycle/sec$

We need to calculate the speed of the wave

Using formula of speed

$v=\lambda\times f$

Put the value into the formula

$v=4.8\times0.625$

$v=3\ m/s$

Hence, The speed of the wave is 3 m/s.

7 0

2 years ago