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FinnZ [79.3K]
3 years ago
13

Just before it is struck by a racket, a tennis ball weighing 0.560 N has a velocity of (20.0m/s)ı^−(4.0m/s)ȷ^(20.0m/s)ı^−(4.0m/s

)ȷ^. During the 3.00 ms that the racket and ball are in contact, the net force on the ball is constant and equal to −(380N)ı^+(110N)ȷ^−(380N)ı^+(110N)ȷ^. What are the x- and y-components (a) of the impulse of the net force applied to the ball; (b) of the final velocity of the ball?
Physics
1 answer:
Art [367]3 years ago
5 0

Answer:

(a) Jx = -1.14Ns, Jy = 110×3×10-³ = 0.330Ns (b) V = (0m/s)ı^−(1.79m/s)ȷ^

Explanation:

Given

W = 0.56N = mg

m = 0.56/g = 0.56/9.8 = 0.057kg

t = 3.00ms = 3.00×10-³s

Impulse is a vector quantity so we would treat it as such

We have been given the force and velocity in their component forms so to get the impulse from these quantities, we pick the respective component for the quantity we want to calculate and do the necessary calculation. The masses are scalar quantities and so do not affect the signs used in the calculations whether positive or negative. So we have that

u = (20.0m/s)ı^−(4.0m/s)ȷ^

ux = 20m/s

uy = – 4.0m/s

F = – (380N)ı^+(110N)ȷ^

Fx = –380N

Fy = 110N

J = impulse = force × time = F×t

So Jx = Fx ×t

Jy = Fy×t

Jx = –380×3×10-³ = -1.14Ns

Jy = 110×3×10-³ = 0.330Ns

Impulse also equals the change in momentum of the body. So

J = m(v–u)

J/m = v – u

V= J/m + u

Vx = Jx/m + ux

Vx = –1.14/0.057 + 20

Vx = -20 + 20 = 0m/s

Vx = 0m/s

Vy= Jy/m + uy

Vy= 0.33/0.057 + (-4.0)

Vy= 5.79 + (-4.0) = 1.79m/s

V = (0m/s)ı^−(1.79m/s)ȷ^

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A certain car's drive-train produces a force of 5300 N as it accelerates from 0
lakkis [162]

The power is 7.1\cdot 10^4 W

Explanation:

First of all, we need to find the acceleration of the car, which is given by

a=\frac{v-u}{t}

where

v = 60 mph = 26.8 m/s is the final velocity

u = 0 is the initial velocity

t = 10.0 s is the time

Substituting,

a=\frac{26.8-0}{10.0}=2.68 m/s^2

Now we can find the mass of the car by using Newton's second law:

F=ma

where

F = 5300 N is the force applied

m is the mass

a=2.68 m/s^2 is the acceleration

Solving for m,

m=\frac{F}{a}=\frac{5300}{2.68}=1978 kg

Now we can use the work-energy theorem, which states that the work done is equal to the change in kinetic energy of the car, to find the work:

W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2

And substituting,

W=\frac{1}{2}(1978)(26.8)^2-0=7.10\cdot 10^5 J

Finally, we can find the power output of the car:

P=\frac{W}{t}

where

W is the work

t = 10.0 s is the time elapsed

Substituting,

P=\frac{7.1\cdot 10^5}{10.0}=7.1\cdot 10^4 W

Learn more about power:

brainly.com/question/7956557

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4 years ago
During which stage of sleep does most dreaming occur
sukhopar [10]
The answer is a rem sleep
7 0
3 years ago
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I need help with 2 questions . Please help c:
dezoksy [38]
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Therefore the correct answer is "all of the above".

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8 0
3 years ago
A cord is wrapped around the rim of a solid uniform wheel 0.300 m in radius and of mass 8.00 kg. A steady horizontal pull of 34.
disa [49]

Answer:

A. α = 94.4 rad/s

B. a = 28.32 m/s

C. N = 34N

D. α = 94.4 rad/s

    a = 28.32 m/s

     N =  44.4  N

Explanation:

part A:

using:

∑T = Iα

where T is the torque, I is the moment of inertia and α is the angular momentum.

firt we will find the moment of inertia I as:

I = \frac{1}{2}MR^2

Where M is the mass and R is the radius of the wheel, then:

I = \frac{1}{2}(8 kg)(0.3 m)^2

I = 0.36 kg*m^2

Replacing on the initial equation and solving for α, we get::

∑T = Iα

Fr = Iα

34 N =  0.36α

α = 94.4 rad/s

part B

we need to use this equation :

a = αr

where a is the aceleration of the cord that has already been pulled off and r is the radius of the wheel, so replacing values, we get:

a = (94.4)(0.3 m)

a = 28.32 m/s

part C

Using the laws of newton, we know that:

N = T

where N is the force that the axle exerts on the wheel part and T is the tension of the cord

so:

N = 34N

part D

The anly answer that change is the answer of the part D, so, aplying laws of newton, it would be:

-Mg + N +T = 0

Then, solving for N, we get:

N = -T+Mg

N = -34 + (8 kg)(9.8)

N =  44.4  N

6 0
3 years ago
Nine tree lights are connected inparallel across 120-V potential difference. The cord to the wall socket carries a current of 0.
jok3333 [9.3K]

Answer:

a)3000ohm

b)4.44mA

Explanation:

a) we were given a Nine tree lights connected inparallel across 120-V potential difference, since the resistor are in parallel we use the expresion below

1/R(total)= 1/R₁ + 1/R₂ + 1/R₃ + 1/R₃ +.... 1/R₉

But according to ohm'law which can be expressed below

V=IR

R=V/I

R(total)= 120/0.36

= 333.33ohm

1/R(total)= 1/R₁ + 1/R₂ + 1/R₃ + 1/R₃ +.... 1/R₉

R₁=R₂ =R₃ =R₄= R₅=R₆=R₇=R₈=R₉

1/R(total)=9/R

1/333.33= 9/R

R= 3000ohm

Therefore, the resistance is 3000ohm

b)the bulbs were connected in series here, then for series connection we use below expression

R₁=R₂ =R₃ =R₄= R₅=R₆=R₇=R₈=R₉

R(total)=9R

= 9*3000

=27000ohm

I=VR

I=V/R

I= 120/27000

= 4.44*10⁻³A

4.44mA

Therefore, the current is 4.4mA

7 0
4 years ago
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