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Marina86 [1]
3 years ago
8

A flea jumps straight up to a maximum height of 0.400 m . what is its initial velocity v0 as it leaves the ground?

Physics
1 answer:
timama [110]3 years ago
8 0

For an object`s motion, the Kinematic equation is,

v^2=v_{0}^2+2ah

Here, v is the final velocity and h is stands for the height of the object and a is the acceleration of the object.

As according to question,

v=0m/s,a=g-9.8 m/s^2 and h = 0.400 m

Thus, putting these values in above equation, we get

0= v_{0}^2 -2gh

or

v_{0} =\sqrt{2 \times 9.8 \times 0.400 }

v_{0} = 2.8 m/s

Therefore, initial velocity is 2.8 m/s



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Gnoma [55]

Answer: 55.52 *10^-6 C= 55.52 μC

Explanation: In order to solve this question we have to take into account the following expressions:

potential energy stired in a capacitor is given by:

U=Q^2/(2*C)  where Q and C are the charge and capacitance of the capacitor.

then we have:

Q^2= 2*C*U=

C=εo*A/d where A and d are the area and separation of the parallel plates capacitor

Q^2=2*εo*A*U/d=2*8.85*10^-12*1.9*10^-5*11*10^3/(1.2*10^-3)=

=55.52 *10^-6C

4 0
2 years ago
Which shared property belongs in the region marked "B"?
NNADVOKAT [17]
The properti that you didnt share
4 0
3 years ago
A car with a mass of 1380 Kg is traveling at 23 m/s to the north. A truck with a mass of 1625 Kg is traveling at 26 m/s to the s
trasher [3.6K]

Answer: -3.49 m/s (to the south)

Explanation:

This problem can be solved by the Conservation of Momentum principle which establishes the initial momentum p_{i} must be equal to the final momentum p_{f}, and taking into account this is aninelastic collision:

Before the collision:

p_{i}=mV_{o}+MU_{o} (1)

After the collision:

p_{f}=(m+M)V_{f} (2)

Where:

m=1380 kg is the mass of the car

V_{o}=23 m/s is the velocity of the car, directed to the north

M=1625 kg is the mass of the truck

U_{o}=-26 m/s is the velocity of the truck, directed to the south

V_{f} is the final velocity of both the car and the truck

p_{i}=p_{f} (3)

mV_{o}+MU_{o}=(m+M)V_{f} (4)

Isolating V_{f}:

V_{f}=\frac{mV_{o}+MU_{o}}{m+M} (5)

V_{f}=\frac{(1380 kg)(23 m/s)+(1625 kg)(-26 m/s)}{1380 kg+1625 kg} (6)

Finally:

V_{f}=-3.49 m/s The negative sign indicates the direction of the velocity is to the south

8 0
3 years ago
It is friction that provides the force for a car to accelerate, so for high-performance cars the factor that limits acceleration
bearhunter [10]

Answer:

About 12 seconds

Explanation:

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3 years ago
A softball player moving 3.89 m/s
Ahat [919]

Answer:

0.119 s

Explanation:

Given that

U=3.89\ m/s\\a=-1.44\ m/s^2\\S=4.8\ m

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We know that

V^2=U^2+2aS\\V^2=3.89^2-2\times 1.44\times 4.8\\V=3.7\ m/s

Also We know that

V=U+at\\

Putting the value of V ,U and a in the previous equation  We get

3.7=3.89-1.44\times t\\t=0.119\ s

Therefore slide time will be 0.119 s

3 0
3 years ago
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