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4 years ago

5

Select the correct equations that show that the gravitational potential energy of a 1000-kg boulder raised 6 m above ground leve

l is 60,000 J. (You can express g in units of N/kg because m/s2 is equivalent to N/kg.)

1 answer:

wel4 years ago

6 0

Answer:

$U = mgh$

Explanation:

The gravitational potential energy of an object is defined as:

$U=mgh$

where

m is the mass of the object

g is the gravitational acceleration

h is the height of the object above a reference level (usually, the ground)

For the boulder in the problem,

m = 1000 kg

g = 10 N/kg

h = 6 m

Substituting,

$U=(1000)(10)(6)=60,000 J$

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El motor de una licuadora gira a 3600 rpm, disminuye su velocidad angular hasta 2000 rpm realizando 120 vueltas. Calcular: a) La

allsm [11]

Answer:

a) α = -65,2 rad/s².

b) t = 2,57 s.

Explanation:

a) La aceleración angular se puede calcular usando la siguiente ecuación:

$\omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta$

En donde:

$\omega_{f}$ : es la velocidad angular final = 2000 rpm = 209,4 rad/s

$\omega_{0}$ : es la velocidad angular inicial = 3600 rpm = 377,0 rad/s

α: es la aceleración angular=?

θ: es el desplazamiento o número de vueltas = 120 rev = 754,0 rad

Las conversiones de unidades se hicieron sabiendo que 1 revolución = 2π radianes y que 1 minuto = 60 segundos.

Resolviendo la ecuación (1) para α, tenemos:

$\alpha = \frac{\omega_{f}^{2} - \omega_{0}^{2}}{2\theta} = \frac{(209,4 rad/s)^{2} - (377,0 rad/s)^{2}}{2*754,0 rad} = -65,2 rad/s^{2}$

Entonces, la aceleración angular es -65,2 rad/s². El signo negativo se debe a que el motor está desacelerando.

b) El tiempo transcurrido se puede encontrar como sigue:

$\omega_{f} = \omega_{0} + \alpha t$

Resolviendo para t, tenemos:

$t = \frac{\omega_{f} - \omega_{0}}{\alpha} = \frac{209,4 rad/s - 377,0 rad/s}{-65,3 rad/s^{2}} = 2,57 s$

Por lo tanto, el tiempo transcurrido fue 2,57 s.

Espero que te sea de utilidad!

3 0

3 years ago

A car hits a wall with a force of 80 N and

ASHA 777 [7]

Formula for acceleration: A=Force/Mass
A=F/M
A=80N/4.5kg
A=17.7 m/s^2

Unit for A is meters per second (m/s^2)
Hope this helps :)

5 0

3 years ago

The moon has a weaker gravitational force than earth. Sofia weighs 50 lbs. on earth. How much will she weigh on the moon? exactl

Neporo4naja [7]

She would weigh less than 50 pounds. everything is super light on the moon. remember that.

3 0

3 years ago

Read 2 more answers

on earth a block is placed on a frictionless table. When a 50 North horizontal force is applied to the block, it accelerates at

melamori03 [73]

As per Newton's II law we know that

$F = ma$

here

F = applied unbalanced force

m = mass of object

a = acceleration of object

now it is given that force F = 50 N North applied on block on earth due to which block will accelerate by 4 m/s^2

so here from above equation

$50 = m* 4$

$m = \frac{50}{4} = 12.5 kg$

Now we took another situation where block is placed on surface of moon and again force F = 25 N is applied on the block

So we will again use Newton's II law

$F = ma$

$25 = 12.5 * a$

$a = \frac{25}{12.5}$

$a = 2 m/s^2$

so block will accelerate on moon by acceleration 2 m/s^2

5 0

3 years ago

During heavy rain, a section of a mountainside measuring 2.3 km horizontally (perpendicular to the slope), 0.75 km up along the

Jobisdone [24]

Answer:

93328 kg

Explanation:

$\rho$ = Density of mud = 1900 kg/m³

Volume of the cuboid is given by

$V=2300\times 750\times 1.4\\\Rightarrow V=2415000\ m^3$

Volume is also given by

$V=Al\\\Rightarrow l=\dfrac{V}{A}\\\Rightarrow l=\dfrac{2415000}{520\times 520}\\\Rightarrow l=8.9312\ m$

The length of the cuboid is 0.89312 m

For the small volume

$V_a=al\\\Rightarrow V_a=5.5\times 8.9312\\\Rightarrow V_a=49.12\ m^3$

Mass is given by

$m=\rho V_a\\\Rightarrow m=1900\times 49.12\\\Rightarrow m=93328\ kg$

The mass of the mud is 93328 kg

6 0

3 years ago