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exis [7]
3 years ago
12

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coas

t in neutral so as to just reach the top of the 150 m high hill at B with vB
Physics
1 answer:
Pavel [41]3 years ago
3 0

Complete question is:

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.

Answer:

(V_A) = 31.32 m/s

Explanation:

We are given;

car's mass, m = 1200 kg

h_A = 100 m

h_B = 150 m

v_B = 0 m/s

From law of conservation of energy,

the distance from point A to B is;

h = 150m - 100 m = 50 m

From Newton's equations of motion;

v² = u² + 2gh

Thus;

(V_B)² = (V_A)² + (-2gh)

(negative next to g because it's going against gravity)

Thus;

(V_B)² = (V_A)² - (2gh)

Plugging in the relevant values;

0² = (V_A)² - 2(9.81 × 50)

(V_A) = √981

(V_A) = 31.32 m/s

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An athlete is working out in the weight room. He steadily holds 50 kilograms above his head for 10 seconds. Which statement is t
postnew [5]

Answer:

A. The athlete isn’t doing any work because he doesn’t move the weight.

Explanation:

We must remember the definition of work, which says that work is equal to the product of mass by the distance displaced. In this case, the athlete only does work when he lifts the weight from the ground to the point where he holds the weight suspended.

So when he's holding the weight, he doesn't do any work.

3 0
3 years ago
10. What is momentum of the water bottle before the collision? (read the
Alona [7]

Answer:

O ksm/s

Explanation:

before collision,

Velocity =0

So,momentum of the bottle before collision=mass ×velocity

=mass×0

=0 kgm/s

8 0
3 years ago
A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out
anzhelika [568]

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

KE_{initial} + PE_{initial} = KE_{final} + PE_{final}

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

0 + mgH = \frac{1}{2}mv^{2} + 0

2gH = v^{2}

v = \sqrt{2\times 9.8\times 5} = 9.89\ m/s

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

mv = (m + m')v'

65.0\times 9.89 = (65.0 + 20.0)v'

v' = 7.56\ m/s

Now, time taken for the fall:

h = \frac{1}{2}gt^{2}

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s

Now, the horizontal distance is given by:

x = v't = 7.56\times 0.638 = 4.823\ m

7 0
3 years ago
Read 2 more answers
What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Yo
insens350 [35]

Full Question

Consider two point charges located on the x axis: one charge, Q1 = -12.0 nC , is located at x1 = -1.705m ; the second charge, Q2 = 36.5 nC, is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Your answer may be positive or negative, depending on the direction of the force.

Answer:

6.79E6 N

Explanation:

Given

Q1 is negative and to the left of Q3 the force will be to the left

Q2 is positive and to the right of Q3 the the force will also be to the left

Net Force is calculated as:

Using Coulomb's law

Coulomb's law: F = kqQ / r²

the constant k = 8.99 x 10^9 N m2 / C2

F = -kQ1*Q3/(r1)² -kQ2*Q3/(r2)²)

F = -kQ3(Q1/(r1)² + Q2/(r2)²)

Where

Q1 = -12nC = -12 * 10^-9C

Q2 = 36.5nC = 36.5 * 10^-9C

Q3 = 49.5nC = 49.5 * 10^-9C

x1 = -1.705m

x2 = x = 0

x3 = -1.170m

r1 = x3 - x1

r1 = -1.170 - -1.705

r1 = -1.170 + 1.705

r1 = 0.535

r1² = 0.286225

r2 = x3

r2 = -1.170

r2² = -1.170²

r2² = 1.3689

So,

F = (-8.99 * 10^9)(49.5 *10^-9) [-12 * 10^-9/0.286225 + 36.5 * 10^-9/1.3689]

F = -445.005 (−4.192505895711E−8 + 2.6663744612462E−8)

F = -445.005 * −1.5261314344648E−8

F = -(8.99 * 10^9) * (49.5 * 10^-9) * [ (-12 * 10^-9) /(-1.770 - -1.705)² + (36.5 * 10^-9)/(-1.170)²]

F = -445.005( −0.000002813572941777538)

F = 0.00000679136118994008324

F = 6.79E6 N

3 0
3 years ago
according to the theory developed by Charles darwin what process causes organisms to develop traita that help them adapt to chan
Anettt [7]

the answer is natural selection

8 0
3 years ago
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