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Nikitich [7]
3 years ago
7

A child bounces a 51 g superball on the sidewalk. The velocity change of the super bowl is from 22 m/s downward to 14 m/s upward

. If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk?
Physics
1 answer:
noname [10]3 years ago
8 0

Answer:

F=1.02x10^{-3} N

Explanation:

From the exercise we know:

m=51g*\frac{1kg}{1000g}=0.051kg

v_{1}=-22m/s

v_{2}=14m/s

t_{2}-t_{1}=1800s

So, the average acceleration is:

a=\frac{v_{2}-v_{1}}{t_{2}-t_{1}}=\frac{(14-(-22))m/s}{1800s}=0.02m/s^2

The average force is:

F=m*a=(0.051kg)(0.02m/s^2)=1.02x10^{-3} N

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With what minimum speed must you toss a 110 g ball straight up to just touch the 11-m-high roof of the gymnasium if you release
Sholpan [36]

Answer:

 v = 13.79 m/s

Explanation:

given,

mass of ball = 110 g

height = 11 m

ball is released from = 1.3 m

minimum speed = ?

using conservation of energy

Potential energy is conserved in the form of kinetic energy

P E =KE

m g h= \dfrac{1}{2}mv^2

v^2 = 2 g h

v=\sqrt{2 g h}

v=\sqrt{2\times 9.8 \times (11-1.3)}

v=\sqrt{2\times 9.8 \times 9.7}

v=\sqrt{190.12}

       v = 13.79 m/s

7 0
3 years ago
a boy throw a ball upward from the top of a cliff 73m high.Calculate the time in which ball will fall on the ground and the velo
LekaFEV [45]

The ball rises for v/g seconds; which equals 14.7/9.8=1.5 seconds . After this time, it’s height will be:


h(t)=g/2(1.5)²+14.7(1.5)


=-4.9 x 2.25 + 22.05


=11.025m


The ball then falls for 49+11.025=60.025m, which takes:


g/2t²=60.025


t²=12.25


t=3.5 secs


Total time: 1.5+3.5=5 seconds

3 0
3 years ago
Two identical 7.10-gg metal spheres (small enough to be treated as particles) are hung from separate 700-mmmm strings attached t
nlexa [21]

Answer:

Explanation:

Let m be mass of each sphere and θ be angle, string makes with vertex in equilibrium.

Let T be tension in the hanging string

T cosθ = mg ( for balancing in vertical direction )

for balancing in horizontal direction

Tsinθ = F ( F is force of repulsion between two charges sphere)

Dividing the two equations

Tanθ = F / mg

tan17 = F / (7.1 x 10⁻³ x 9.8)

F = 21.27 x 10⁻³ N

if q be charge on each sphere , force of repulsion between the two

F = k q x q / r² ( r is distance between two sphere , r = 2 x .7 x sin17  = .41 m )

21.27 x 10⁻³  = (9 X 10⁹ x q²) / .41²

q² = .3973 x 10⁻¹²

q = .63 x 10⁻⁶ C

no of electrons required  = q / charge on a single electron

= .63 x 10⁻⁶ / 1.6 x 10⁻¹⁹

= .39375 x 10¹³

3.9375 x 10¹² .

4 0
3 years ago
I need 1, 2 and 3 <br><br> Please help!
svetoff [14.1K]

1)Kenetic Energy is defined as energy which a body possesses by virtue of being in motion. 2)KE) is KE = 0.5 x mv2. Here m stands for mass, the measure of how much matter is in an object, and v stands for the velocity of the object, or the rate at which the object changes its position..

And I hope this helped :)

7 0
3 years ago
A family drives north for 30km then turns east for 20km. The family then decided to turn west for 5km before finally stopping to
salantis [7]

Answer:

A

Explanation:

They drove 30km north. The displacement adds up to 25km therefore making the distance greater

Hope this helps!

6 0
3 years ago
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