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2 years ago

An inclined plane is used to lift a box into a moving truck. The ramp is 6m long and 1.5m

Physics

1 answer:

ElenaW [278]2 years ago

5 0

Given :

An inclined plane is used to lift a box into a moving truck. The ramp is 6 m long and 1.5 m.

To Find :

The mechanical advantage of the inclined plane.

Solution :

Mechanical advantage, M.A of an inclined plane is given by :

$M.A = \dfrac{\text{Length of slope}}{\text{Height of plane}}$

Putting given values in above equation, we get :

$M.A = \dfrac{6 \ m}{1.5 \ m}\\\\M.A = 4$

Therefore, the mechanical advantage of the inclined plane is 4.

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On topographic maps, contour lines that are farther apart indicate what ?

gladu [14]

Answer:

if I am correct, they indicate less steep terrain. think of it as the steeper the terrain the closer together the lines would be. hope that makes sense for you guys.

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2 years ago

A mass of 1.00×10−6 μkg is the same as

tankabanditka [31]

I'm not sure but I know u is 10^6

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2 years ago

Wilma can mow a lawn in 80 minutes. Rocky can mow the same lawn in 120 minutes. Construct an equation that would allow you to de

xeze [42]

Answer:

$t = 96 minutes$

Explanation:

Time to mow 1 lawn by Wilma is 80 minutes

so work done in 1 minute by Wilma is given as

$W_1 = \frac{1}{80}$

Similarly Rocky mow same lawn in 120 minute

so work done in 1 minute by Rocky is given as

$W_2 = \frac{1}{120}$

now we know that they both worked by "t" time

so total work performed by them

$W = \frac{t}{80} + \frac{t}{120}$

they both mow 2 lawns then it is given as

$2 = (\frac{1}{80} + \frac{1}{120})t$

$t = 96 minutes$

5 0

2 years ago

A 20-kg barrel is rolled up a 20-m ramp to the back of a truck whose floor is 5.0 m above the ground. What work is done in loadi

oee [108]

The angle of inclination is calculated using sin function,

sin θ = 5 m / 20 m = 0.25

θ = 14.4775°

The net force exerted is then calculated:
F net = m g sin θ = 20 * 9.8 * 0.25

F net = 49N

Work is product of net force and distance:
W = F net * d = 49 * 20

Work = 980 J

4 0

2 years ago

A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calcula

Luda [366]

Answer:

a) 0.1832 A

b) 11.91 Volts

c) 2.18 Watt , 0.0168 Watt

Explanation:

(a)

R = external resistor connected to the terminals of the battery = 65 Ω

E = Emf of the battery = 12.0 Volts

r = internal resistance of the battery = 0.5 Ω

i = current flowing in the circuit

Using ohm's law

E = i (R + r)

12 = i (65 + 0.5)

i = 0.1832 A

(b)

Terminal voltage is given as

$V_{ab}$ = i R

$V_{ab}$ = (0.1832) (65)

$V_{ab}$ = 11.91 Volts

(c)

Power dissipated in the resister R is given as

$P_{R}$ = i²R

$P_{R}$ = (0.1832)²(65)

$P_{R}$ = 2.18 Watt

Power dissipated in the internal resistance is given as

$P_{r}$ = i²r

$P_{r}$ = (0.1832)²(0.5)

$P_{r}$ = 0.0168 Watt

5 0

2 years ago