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Readme [11.4K]
3 years ago
5

If a bat with a mass of 5 kg and acceleration of 2 m/s2 hits a ball whose mass is 0.5 kg in the forward direction, what is the r

eaction force of the ball on the bat?
10 N, forward
0 N, forward
0.5 N, upward
10 N, backward
Physics
2 answers:
ivanzaharov [21]3 years ago
8 0

Answer:

10 N , forward

Explanation:

F=m×a

F= 5×2

F=10 N

lina2011 [118]3 years ago
6 0

Answer:

10 N, backward

Explanation:

Force exerted on the ball by the bat can be calculated according to newtons second law,

Force = mass × acceleration

Given mass of the ball = 5kg

Acceleration of the ball = 2m/s²

Force exerted on the ball= 5×2

Force exerted on the ball by the bat = 10N

If the force of 10N is applied on the ball by the bat in the forward direction, the reaction force of the ball on the bat will also be 10N but in the opposite direction (backward) to the direction of the bat according to newton's third law which states that actions and reaction are equal and opposite.

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A horizontal line labeled B has an arrow labeled A strike it from right and above and then another arrow D emerges from the stri
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Answer:

c is the actual answer.

Explanation:

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Write the equation expressing the relationship "y varies directly as x." use k as the constant of proportionality.
Alexeev081 [22]

The equation expressing the statement "y varies directly as x"  is y=kx.

Explanation

The statement that y varies directly as x is analogous to saying that  the ratio of y to x is constant. In other words, when x increases, y likewise increases and that when x decreases, y decreases proportionally.

Mathematically, we express the relationship that the ration of y is to x is constant is expressed as; \frac{y}{x} =k where k is the constant of proportionality.

We can then solve the relationship for y to determine the correct form of the relationship as shown below,

\frac{y}{x} =k\\\rightarrow y=kx


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3 years ago
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A body weights 50 N in air and 45 N when wholly immersed in water calculate (i) the loss in weight of the body in water (ii) the
Lelechka [254]

Answer:

difference  \: in \: weight = 150n - 100n = 50n

Now,buyantant force

difference \: in \: weight \: = volume(body) \times density \: of \: water \:  \times g

so;

50 =  {v}^{b}  \times 1 \times  {10}^{3}  \times 9.8m {s}^{2}

{v}^{b}  =  \frac{50}{1000 } \times 9.8

=  \frac{50}{9800}

= 0.0051

Now,

mass \: in \: air \:  = 150n =  \frac{150}{9.8kg}

density =  \frac{weght}{volume}

=  \frac{150}{0.0051}  \times 9.8 \\ x = 3000

And now,

specific \: density \:  =  \frac{density of \: the \: body}{density \: of \: water}

=  \frac{3000}{1000}

= 3

Hence that,specific density of a given body is 3

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3 0
3 years ago
An object dropped from rest from the top of a tall building on planet x falls a distance d(t)18 left parenthesis t right parenth
melamori03 [73]

displacement is given by equation

d = 18t^2

now at t = 5 s the position is

d_1 = 18 *5^2 = 450 m

similarly position at t = 9 s

d_2 = 18*9^2 = 1458 m

so the displacement of object in given interval of time will be

d = 1458 - 450 = 1008 m

time interval

\delta t = 9 - 5 = 4 s

now the average velocity will be given as

v = \frac{\delta x}{\delta t}

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4 0
3 years ago
The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 c
Reika [66]

Answer:

the knee extensors must exert 15.87 N

Explanation:

Given the data in the question;

mass m = 4.5 kg

radius of gyration k = 23 cm = 0.23 m

angle ∅ = 30°

∝ = 1 rad/s²

distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m

using the expression;

ζ = I∝

ζ = mk²∝

we substitute

ζ = 4.5 × (0.23)² × 1

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so

from; ζ = rFsin∅

F = ζ / rsin∅

we substitute

F = 0.23805 / (0.03 × sin( 30 ° )

F = 0.23805 / (0.03 × 0.5)

F F = 0.23805 / 0.015

F = 15.87 N

Therefore, the knee extensors must exert 15.87 N

7 0
3 years ago
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