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3 years ago

you push the ball with a force of 22.8N which induces a -2.3 frictional force what is the net force while you push the ball

1 answer:

4 0

Force is a vector quantity in which direction matters. This means that the forces will act in opposing directions so the net force will be
22.8 - 2.3 =
20.5 Newtons

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Sam heaves a 16-lb shot straight up, giving it a constant upward acceleration from rest of 35.0 m/s2 for 64.0 cm. He releases it

stich3 [128]

Answer:

a. 6.69m/s

b. y=4.48m

c. t=1.43secs

Explanation:

Data given, acceleration,a=35m/s^2

distance covered,d=64cm=0.64m,

a. to determine the speed, we use the equation of motion

initial velocity,u=0m/s

if we substitute values we arrive at

$v^{2}=u^{2}+2as\\v^{2}=0+2*35*0.64\\v^{2}=44.8m/s\\v=\sqrt{44.8}\\ v=6.69m/s\\$

b. After taking the shot,the acceleration value is due to gravity i.e a=9.81m/s^2

and the distance becomes (y-2.2) above the ground. When it reaches the maximum height, the final velocity becomes zero and the initial velocity becomes 6.69m/s.

Hence we can write the equation above again

$v^{2}=u^{2}-2a(y-2.2)\\$

if we substitute values we have

$v^{2}=u^{2}-2a(y-2.2)\\0=6.69^{2}-2*9.81(y-2.2)\\y-2.2=\frac{44.76}{19.62} \\y=2.28+2.2\\y=4.48m$

c. the time it takes to arrive at 1.83m is obtain by using the equation below

$1.83-2.2=6.69t-\frac{1}{2} *9.81t^{2}\\4.9t^{2}-6.69t-0.37\\using \\t= \frac{-b±\sqrt{b^{2}-4ac} }{2a}\\ where \\a=4.9, b=-6.69, c=-0.37$

if we insert the values, we solve for t , hence t=1.43secs

6 0

3 years ago

Which statement is true of convex mirrors?

Leno4ka [110]

I think it is d if not then im sorry

7 0

3 years ago

You are in downtown Chicago (where streets run N-S and E-W). You started from 600 N. Michigan Avenue, and walked 3 blocks toward

pickupchik [31]

Answer:

Displacement is 565.69 m at 45° west of north

Explanation:

Let north represent positive y axis and east represent positive x axis.

We have journey started from 600 N. Michigan Avenue, and walked 3 blocks toward north, 4 blocks toward west, and 1 block toward north to a train station.

3 blocks toward north = 300 j m

4 blocks toward west = -400 i m

1 blocks toward north = 100 j m

Total displacement = -400 i + 400 j m

Magnitude

$s=\sqrt{(-400)^2+400^2}=565.69m$

Direction,

$\theta =tan^{-1}\left ( \frac{400}{-400}\right )=135^0$

Direction is 45° west of north.

Displacement is 565.69 m at 45° west of north

4 0

2 years ago

On a horizontal frictionless floor, a worker of weight 0.900 kN pushes horizontally with a force of 0.200 kN on a box weighing 1

Ad libitum [116K]

Answer:

D) The worker will accelerate at 2.17 m/s² and the box will accelerate at 1.08 m/s² , but in opposite directions.

Explanation:

Newton's third law

Newton's third law or principle of action and reaction states that when two interaction bodies appear equal forces and opposite directions. in each of them.

F₁₂= -F₂₁

F₁₂: Force of the box on the worker

F₂₁: Force of the worker on the box

Newton's second law

∑F = m*a

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Formula to calculate the mass (m)

m = W/g

Where:

W : Weight (N)

g : acceleration due to gravity (m/s²)

Data

W₁ =1.8 kN : box weight

W₂ = 0.900 kN : worker weight

F₂₁ = 0.200 kN

F₁₂ = - 0.200 kN

g = 9.8 m/s²

Newton's second law for the box

∑F = m*a

F₂₁ = m₁*a₁ m₁=W₁/g

0.2 kN = (1.8kN)/(9.8 m/s² ) *a₁

$a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{1.8 kN}$

a₁= 1.08 m/s² : acceleration of the box

Newton's second law for the worker

∑F = m*a

F₁₂ = m₂*a₂ , m₂=W₂/g

- 0.2 kN =( (0.9 kN) /(9.8 m/s² ) )*a₂

$a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{0.9 kN}$

a₂= -2.17 m/s² : acceleration of the worker

5 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

xxMikexx [17]

14-6 =8
8/4= 2m/s per second

8 0

3 years ago

Read 3 more answers