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Yanka [14]
3 years ago
13

a tiger leaps horizontally from a 5.7m high rock with a speed of 4.1m/s. How far from the base of the rock will she land

Physics
1 answer:
shepuryov [24]3 years ago
6 0
Looks like a projectiles problem, with the tiger as the projectile. See enclosed for guidance.For more information/help, please ask.ps there's a poem which contains the lines "Tyger tyger burning bright, in the forest of the night; what immortal hand or eye; could frame thy fearful symmetry ?"

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Suppose that two objects attract each other with a gravitational force of 16 units. If the distance between the two objects is r
Naddik [55]

Explanation:

Fgravity = G*(mass1*mass2)/D²

G is the gravitational constant throughout the universe.

D is the distance between both objects.

D is now reduced by a factor of 5, meaning Dnew = D/5 we get

Fgravitynew = G*(mass1*mass2)/(D/5)² =

= G*(mass1*mass2)/(D²/25) =

= 25* G*(mass1*mass2)/D² = 25* Fgravity

the new force of gravity/attraction is 25×16 = 400 units.

4 0
2 years ago
The block slides on a horizontal frictionless surface. The block has a mass of 1.0kg and is pushed 5.0N at 45°. What is the magn
Firlakuza [10]

Answer:

3.5\:\mathrm{m/s^2}

Explanation:

Newton's 2nd law is given as \Sigma F = ma.

To find the acceleration in the horizontal direction, you need the horizontal component of the force being applied.

Using trigonometry to find the horizontal component of the force:

\cos 45^{\circ}=\frac{x}{5},\\\frac{\sqrt{2}}{{2}}=\frac{x}{5},\\x=\frac{5\sqrt{2}}{2}

Use this horizontal component of the force to solve for for the acceleration of the object:

\frac{5\sqrt{2}}{2}=1.0\cdot a,\\a=\frac{5\sqrt{2}}{2}\approx \boxed{3.5\:\mathrm{m/s^2}}

8 0
2 years ago
a 2,000 pound car is driving at 60 miles/hour along a straight, level road. what is the net force acting on the car?
SVETLANKA909090 [29]

Answer:

0

Explanation:

According to Newton's second law, the net force is equal to the mass times the acceleration.  Since the car is not accelerating, the net force is 0.

5 0
3 years ago
A crate is lifted vertically 1.5 m and then held at rest. The crate has weight 100 N (i.e., it is supported by an upward force o
bogdanovich [222]

Answer:

1) W = 150 J

Explanation:

Work (W) is defined as the product of force F by the distance (d)the body travels due to this force.  

W= F*d Formula ( 1)

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

The component of the force that performs work must be parallel to the displacement.  

Work done to lift the floor box to its final position

We apply the formula (1)

W= F*d

W = (100 N)*(1.5 m)

W = 150 J

5 0
3 years ago
The fully loaded bus accelerates uniformly from rest to a speed of 14 m / s. The time taken to reach a speed of 14 m / s is 20 s
tino4ka555 [31]

Answer:

0.7m/s^2

Explanation:

acceleration=(final-initial velocity)/time

x=(14-0)/20

x=14/20

x=0.7

3 0
2 years ago
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