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3 years ago

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The (nonconservative) force propelling a 1.50 × 103-kg car up a mountain road does 4.70 × 106 J of work on the car. The car star

ts from rest at sea level and has a speed of 27.0 m/s at an altitude of 2.00 × 102 m above sea level. Obtain the work done on the car by the combined forces of friction and air resistance, both of which are non conservative forces

1 answer:

aliina [53]3 years ago

6 0

Answer:

be the mehcanic for once and see what the answer is

Explanation:

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a child's toy consists of a piece of plastic attached to a spring. the spring is compressed against the floor a distance of 2.0

Nat2105 [25]

Answer:

1.7N

Explanation:

Force = kx

Where x = spring compression and

K = spring constant

K =85N/m

x = 2.0cm / 100

= 0.02m

Force = 85 x 0.02

= 1.7N

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3 years ago

Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an

ddd [48]

Answer:

Al's mass is 102.92 kg

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

$F_{net} = 0$

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

$F_{net} = \frac{dp_{horizontal}}{dt} = 0$

$p_{horizontal_i }= p_{horizontal_f}$

where the suffix i and f means initial and final respectively.

The initial momentum will be:

$p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}$

But, as they are at rest, initially

$p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0$

$p_{horizontal_i} = 0$

So, this means:

$p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0$

We know that the have an combined mass of 195 kg:

$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

so:

$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

$v_{Jo_f}= - 11.4 \frac{m}{s}$

where the minus sign appears as they are moving at opposite directions

$m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }$

$m_{Al} = 102.92 \ kg$

and this is the Al's mass.

5 0

3 years ago

A 100 V battery is connected across a

Sergio039 [100]

1 mA = 0.001 A

Therefore, 5 mA = 0.001 * 5

=0.005 A

Resistance = voltage / current

= 100 / 0.005

= 20000 ohms

Current = voltage / resistance

= 25 / 20000

= 0.00125 A (or) 1.25 mA

5 0

3 years ago

A metal can containing condensed mushroom soup has mass 215 g, height 10.8 cm, and diameter 6.38 cm. It is placed at rest on its

s344n2d4d5 [400]

Answer:

Part a)

Moment of inertia of the cylinder is given as

$I = 1.21 \times 10^{-4} kg m^2$

Part B)

Height of the cylinder is of no use here to calculate the inertia

Part C)

Since we don't know about the viscosity data of the soup inside the cylinder so we can't say directly about the moment of inertia of the cylinder as $I = 1/2 mR^2$

Explanation:

As we know that the inclined plane is of length L = 3 m

and its inclination is given as 25 degree

so we know that acceleration of center of mass of the cylinder is constant so we will have

$v_f^2 = v_i^2 + 2 a L$

so we have

$v_f^2 = 0 + 2a(3)$

now we know that

$v_{avg} = \frac{L}{t} = \frac{v_f + v_i}{2}$

$\frac{3}{1.50} = \frac{v_f + 0}{2}$

$v_f = 4 m/s$

Now we have know that final speed of the cylinder due to pure rolling is given as

$v_f = \sqrt{\frac{2gH}{1 + \frac{I}{mR^2}}}$

$4 = \sqrt{\frac{2(9.81)(3 sin25)}{1 + \frac{I}{0.215(0.0319)^2}}}$

$I = 1.21 \times 10^[-4} kg m^2$

Part B)

Height of the cylinder is of no use here to calculate the inertia

Part C)

Since we don't know about the viscosity data of the soup inside the cylinder so we can't say directly about the moment of inertia of the cylinder as $I = 1/2 mR^2$

8 0

3 years ago