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3 years ago

A block of mass m begins at rest at the top of a ramp at elevation h with whatever pe is associated with that height.

2 answers:

8 0

<span>PE = mass * height * g = 87.60 J

Frictional force = m * g * sin theta * uk = 4.7 kg * 9.81 m/sec^2 * sin 36.87 * 0.3 = 8.299 N

Work done by friction = force * distance = 41.50 J

Remaining energy to be converted to KE = 87.60 J - 41.50 J = 41.10 J</span>hope it helps

Darina [25.2K]3 years ago

8 0

Answer:

The PE associated with that height is given as, m×g×h

Explanation:

Potential energy (PE) is the energy possessed by an object when at rest. Its opposite is the kinetic energy (KE). When an object is at a certain height, it would possess potential energy because initially it is at rest. Also the value of the energy is determined by its mass, height and the force of gravity acting on it at that height.

So, the potential energy (PE) = mass× height× value of the gravitational force

= m× g× h

Its unit is Joules. The potential energy of the mass would be converted to kinetic energy when it is in motion.

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The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. If HST has a tangential speed of 7,750 m/s, how long is HS

deff fn [24]

Answer: 5,640 s (94 minutes)

Explanation:

the tangential speed of the HST is given by

$v=\frac{2\pi r}{T}$ (1)

where

$2\pi r$ is the length of the orbit

r is the radius of the orbit

T is the orbital period

In our problem, we know the tangential speed: $v=7,750 m/s$ . The radius of the orbit is the sum of the Earth's radius and the distance of the HST above Earth's surface:

$r=6.38\cdot 10^6 m+569,000 m=6.95\cdot 10^6 m$

So, we can re-arrange equation (1) to find the orbital period:

$T=\frac{2 \pi r}{v}=\frac{2 \pi (6.95\cdot 10^6 m/s)}{7,750 m/s}=5,640 s$

Dividing by 60, we get that this time corresponds to 94 minutes.

6 0

3 years ago

Read 2 more answers

A school bus has a mass (including the driver and passengers) of 1.64 times 10^4 kg and is driving north at a speed of 15.2 km/h

Butoxors [25]

Answer:

Explanation:

Given

mass of bus along with travelers travelling in North direction is $m_1=1.6\times 10^4 kg$

speed of bus towards North $v_1=15.2 km/h\approx 4.22\ m/s$

mass of bus travelling in South direction is $m_2=1.578\times 10^4 kg$

speed of bus $v_2=12.2 km/h\approx 3.38\ m/s$

mass of each Passenger in south moving bus $m_0=64.8 kg$

Momentum of North moving bus

$P_1=m_1\times v_1$

$P_1=1.6\times 10^4\times 4.22$

$P_1=6.768\times 10^4 kg-m/s$

Momentum with south moving bus

$P_2=m_2\times v_2+n\cdot m_0\times v_2$

$P_2=(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38$

For total momentum to be towards south

$P_2-P_1$ should be greater than 0

thus for least value of n

$P_2=P_1$

$(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38=6.768\times 10^4$

$1.578\times 10^4+n\cdot 64.8=2.0023\times 10^4$

$n=\frac{4243.6686}{64.8}=65.48\approx 66$

5 0

3 years ago

this is a 3 part questionOn vacation, your 1400-kg car pulls a 560-kg trailer away from a stoplight with an acceleration of 1.85

mamaluj [8]

ANSWER:

(a) 1036 N

(b) -1036 N

STEP-BY-STEP EXPLANATION:

Given:

Mc = 1400 kg

Mt = 560 kg

a = 1.85 m/s^2

(a)

Force by car on trailer:

$\begin{gathered} F_c=m\cdot a \\ F_c=560\cdot1.85 \\ F_c=1036\text{ N} \end{gathered}$

(b)

$\begin{gathered} F_t=-F_c \\ F_t=-1036\text{ N} \end{gathered}$

(c)

$\begin{gathered} F_n=1400\cdot1.85 \\ F_n=2590\text{ N} \end{gathered}$

3 0

1 year ago

Un coche inicia un viaje de 450 km a las ocho de la mañana con una velocidad media de 90 km/h. ¿A qué hora llegará a su destino?

Artyom0805 [142]

Answer:

Llegara a su destino a la 1:00 pm

Explanation:

Si el coche va a 90 km/h buscamos un numero q al multiplicarlo por 90 nos de 450. Entonces 90×5 = 450, si hacemos la cuenta desde las ocho de la mañana mas las 5 horas del viaje terminaria llegando a su destino a la 1:00 pm.

5 0

2 years ago

Technician A says that the starter motor used to crank diesel engines can draw up to 400 amps of current. Technician B says that

Aleks04 [339]

Answer: Option A : Technician A

Explanation:

The statement/observation, "that the starter motor used to crank diesel engines can draw up to 400 amps of current" made by Technician A is correct.

A diesel engine uses up to 400+ Amperes of electricity to start up a diesel engine in the ignition chamber of motor engine.

8 0

3 years ago