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11111nata11111 [884]
2 years ago
15

"A boat that can travel at 4.0 km/h in still water crosses a river with a current of 2.0 km/h. At what angle must the boat be po

inted upstream (that is, relative to its actual path) to go straight across the river?

Physics
1 answer:
IRISSAK [1]2 years ago
8 0

Answer:

30 degrees

Explanation:

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A cart of mass m moving right at speed v with respect to the track collides with a cart of mass 0.7m moving left.
Zolol [24]

Answer:

10v / 7

Explanation:

Using the conservation law of momentum

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

m v - 0.7 m v₁ =  ( 0.7 m + m) 0 m/s since the cart stuck together after collision. taken right to be positive and left to be negative

m v - 0.7 m v₁ = 0

- 0.7 m v₁ = -m v

v₁ = -m v / - 0.7 m = 10v / 7

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How does a rollercoaster convert potential energy into kinetic energy and then back to potential energy again and again?
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When a roller coster goes down hill all its stored energy is bieng used as electricity is bieng converted into energy (moving)
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How to increase the current in a circuit
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Answer:

So to increase current of the circuit what you can do is :

1. Use conductor of low resistivity, ¶.

2. Use conductor of small length.

3. Use thick wire.

4. Decrease the temperature of the circuit.

5. If operating temprature is high than use semiconductor, because it have negative temprature coefficient.

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A person pulls a crate of mass M = 63 kg a distance 40.0 m along a horizontal floor by a constant force FP = 130 N, which acts a
GrogVix [38]

Answer:

Check attachment for solution and diagrams

Explanation:

Given that,

Mass of crate m=63kg

Distance travelled d=40m

Horizontal force Fx=130N

Angle the force applied on cord makes with horizontal is θ=23°.

The weight of the crate is given by

W=mg

W=63×9.81

W=618.03N

Horizontal force Fx=130N

Resolving the applied Force F to the horizontal will give

Fx=FCos θ

F=Fx/Cos θ

F=130/Cos23

F=141.2N

a. Check attachment for model diagram

b. Check attachment for free body diagram

c. Check attachment for pictorial representation

d. Work done by gravitational force.

We, know that the body did not move upward, then the distance d=0

Work done is given as

W=F×d

So, d=0

W=F×0

W=0J

So, no work is done by gravity

e. Normal force?

Using newton law of motion

ΣFy = may

Since the body is not moving upward, then ay=0m/s²

N+141.2Sin23-618.03=0

N=618.03-141.2Sin23

N=562.86N

f. Work done by normal force.

The body is not moving upward, then the distance is zero

d=0

Work done by normal=normal force × distance

Wn=562.86×0

Wn=0J

No work is done by the normal force

g. Frictional force?

Since the coefficient of kinetic friction is zero, then the surface is frictionless

So, no frictional force is acting on the body

Fictional force is given as

Fr=μk•N

Given that, μk=0

Fr=0×562.86

Fr=0N

d. Work done by frictional force?

Since the frictional force Is zero, then, no work is done by friction

W(friction ) = frictional force × d

Here, the body moved a distance of 40m

W(fr)=0×40

W(fr)=0J

No work is done by friction

I. Work done by exerted force

The horizontal component of the exerted force is 130N and the body traveled a distance of 40m

Then, work done is given as

Workdone=force ×distance

Work done=130×40

W=5200J

W=5.2KJ

h. Net workdone?

Since no work is lost by friction, then, the net work done is equal to the work done by the exerted force.

Went, = work done by force exerted - work done by friction

Wnet=5200-0

Wnet, =5200

Wnet=5.2KJ

7 0
2 years ago
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