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algol [13]
3 years ago
7

A naughty squirrel jumps from a tree limb onto a birdfeeder. The tree limb is 4.5 m above the ground and the top of the birdfeed

er is 1.5 m above the ground. If the squirrel jumps horizontally with a velocity of 2.5 m/s then how far away in the horizontal direction can the bird feeder be for the squirrel to just make it onto the top of the feeder.
Physics
1 answer:
andreev551 [17]3 years ago
7 0

Answer:

Distance of 1.956 m

Explanation:

Consider the vertical plane

S=ut+\frac{1}{2} at^2  

where u=o, S=4.5-15=3 m

3=0\times t+\frac{1}{2} 9.8 t^2\\t=\sqrt{0.612} \\t=0.782 s

now consider horizontal plane

u=2.5 m/s, a=0

S=2.5\times 0.782+0\\S=1.956 m

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Answer:True

Explanation:

Buoyant force is the net upward force, that affect on the object in a fluid

4 0
3 years ago
NEED HELP ASAP
Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum p_{o} must be equal to the final momentum p_{f}:  

p_{o}=p_{f} (1)  

Where:  

p_{o}=m_{1} V_{o} + m_{2} U_{o} (2)  

p_{f}=(m_{1} + m_{2}) V_{f} (3)

m_{1}=110 kg is the mass of the first football player

V{o}=-7 m/s is the velocity of the first football player (to the south)

m_{2}=75 kg  is the mass of the second football player

U_{o}=4 m/s is the velocity of the second football player (to the north)

V_{f} is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f} (4)  

Isolating V_{f}:

V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}} (5)

V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg} (6)

V_{f}=-2.54 m/s (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy \Delta K is defined as:

\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2} (8)

Simplifying:

\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2}) (9)

\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2}) (10)

Finally:

\Delta K=-2351.25 J (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

6 0
3 years ago
Calculate the acceleration of a 1000 kg car if the motor provides a small thrust of 1000 N and the static and dynamic friction c
grin007 [14]

Explanation :

It is given that,

Mass of the car, m = 1000 kg              

Force applied by the motor, F_A=1000\ N

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\dfrac{F_A-\mu mg}{m}=a

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a=-3.905\ m/s^2

So, the acceleration of the car is -3.905\ m/s^2. Hence, this is the required solution.

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Answer:

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Explanation:

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