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Monica [59]
3 years ago
7

It was once recorded that a Jaguar

Physics
1 answer:
Artyom0805 [142]3 years ago
6 0

Answer:

71.85 m/s

Explanation:

Given the following :

Length of skid marks left by jaguar (s) = 290 m

Skidding Acceleration (a) = - 8.90m/s²

Final velocity of jaguar (v) = 0

Speed of Jaguar before it Began to skid =?

Hence, initial speed of jaguar could be obtained using the formula :

v² = u² + 2as

Where

v = final speed of jaguar ; u = initial speed of jaguar(before it Began to skid) ; a = acceleration of jaguar ; s = distance /length of skid marks left by jaguar

0² = u² + (2 × (-8.90) × 290)

0 = u² + (-5,162)

u² = 5162

Take the square root of both sides

u = √5162

u = 71.847 m/s

u = 71.85m/s

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A pilot, whose mass is 84.0 kg, makes a loop-the-loop in a fast jet. Assume that the jet maintains a constant speed of 345 m/s a
arsen [322]

Answer:

The apparent weight is 5 times greater than the original weight at the bottom.

Explanation:

Given:

Mass of the pilot, m = 84 kg

Velocity of the jet, v = 345 m/s

Radius of the loop, R = 3.033 km = 3.033 * 10^3 m

We have to find the apparent weight that the pilot feels.

Let the apparent weight be "N" .

Apparent weight :

  • It is based on where is the position of the pilot in the loop-the-loop.
  • The apparent weight is the highest at the bottom of the loop-the-loop.
  • Because the weight acts down and the normal force acts towards the center of the circle.

From the FBD shown we can say that :

apparent weight (N)

⇒ N=mg+\frac{mv^2}{R}

⇒ N=mg(1+\frac{v^2}{Rg} )

⇒ N=mg(1+\frac{(345)^2}{3.033\times 10^3 \times 9.8} )

⇒ N=5mg

Therefore,

The force exerted by the seat on the pilot at the bottom of the loop is greater than the pilots weight by a factor of 5.

7 0
3 years ago
Suppose the pucks start spinning after the collision, whereas they were not before. Will this affect your momentum conservation
gogolik [260]

Answer:

No, it will not affect the results.

Explanation:

For elastic collisions in an isolated system, when a collision occurs, it means that the systems objects total momentum will be conserved under the condition that there will be no net external forces that act upon the objects.

What that means is that if the pucks start spinning after the collision, we are not told that there was any net external force acting on the puck and thus momentum will be conserved because momentum before collision will be equal to the momentum after the collision.

3 0
3 years ago
Please help im on a timer!! 50 points!!
Crank

Strength of the magnetic field: C) 20 T

Explanation:

When a conductive wire moves in a region with magnetic field, an electromotive force is induced in the wire due to the phenomenon of electromagnetic induction.

If the wire moves perpendicular to the field, the magnitude of the induced emf in the wire is given by

\epsilon=BLv

where

B is the strength of the magnetic field

L is the length of the wire

v is the speed of the wire

For the wire in this problem, we have:

\epsilon=12 V is the emf induced in the wire

v = 3.0 m/s is the speed of the wire

L = 0.20 m is its length

Solving for B, we find the strength of the magnetic field:

B=\frac{\epsilon}{Lv}=\frac{12}{(0.20)(3.0)}=20 T

Learn more about magnetic fields:

brainly.com/question/3874443

brainly.com/question/4240735

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5 0
4 years ago
give an example of how the law of inertia is demonstrated (a) for moving objects and (b) for objects at rest
lakkis [162]

Answer:

Both

Explanation:

Both of them are part of newtons laws of motion

6 0
2 years ago
You do a physics lab experiment on another planet. A small block is released from rest at the top of a long frictionless ramp th
Alex Ar [27]

Answer:

g= 1.8 m/s²

Explanation:

From conservation of energy we know:

mgh=\frac{1}{2}mv^2

Solving for g with h being the height of the ramp with angle ∅ and length L:

(1) g=\frac{v^2}{2h}=\frac{v^2}{2Lsin\theta}

The velocity v at the bottom of the ramp for a constant acceleration a in a time t is given by:

(2)v=at=gsin\theta t

Plugging equation 2 in to equation 1:

(3) g=\frac{2L}{sin\theta t^2}

5 0
4 years ago
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