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4 years ago

A bat hits a 0.150 kg baseball for

Physics

1 answer:

lara31 [8.8K]4 years ago

3 0

Answer:

11.85 kg m/s

Explanation:

impulse = mass ( change in velocity )

= mass ( final velocity - initial velocity )

= 0.150 ( 32.0 - (-47.0))

= 0.150 ( 32.0 +47.0)

= 0.150 (79)

= 11.85 kg m/s

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A luggage handler pulls a 20.0 kgkg suitcase up a ramp inclined at 32.0 ∘∘ above the horizontal by a force F⃗ F→ of magnitude 16

Nuetrik [128]

Answer:

a) W₁ = 8242.2 J, b) W₁ = 8242.2 J , c) W₃ = 0 , d) W₄ = -189.51 J ,

f) v = 27.24 m / s

Explanation:

a) Work is defined by

W = F. d = F d sin θ

where angle is between force and displacement

n this case the suitcase is going up and the outside F is parallel to the plane, so the angle is zero and the cosine is 1

W = F d

Let's calculate

W = 169 3.8

W₁ = 8242.2 J

b) the gravitational force is vertical so it has an angle with respect to the horizontal parallel to the plane of

θ’= 90 - θ

θ'= 90-32 = 58º

W = m g d thing θ ’

W = 20 9.8 3.8 thing (180 + tea ’) =

W = 744.8 cos (180 + 32)

W₂ = -631.6 J

c) The normal work, as it has 90º with respect to the displacement, its work is zero

W₃ = 0

d) the work of the friction force

Let's write Newton's second law the Y axis

N- Wy = 0

Cos 32 = Wy / W

N = W cos 32

The expression for friction force is

fr = μ N

fr = μ mg cos 32

fr = 0.300 20 9.8 cos (32)

fr = 49.87 N

The work of the friction force

W = fr d cos 180

W₄ = -49.87 3.8

W₄ = -189.51 J

E) The total work

W = W₁ + W₂ + W₃ + W₄

W = 8242.2- 631.6 + 0 -189.51

W_total = 7421.09 J

F) Usmeosel theorem of work and energy

W = ΔK

W = ΔK = ½ m v² - 0

v =√ 2W / m

v = √ (2 7421.09 / 20)

v = 27.24 m / s

5 0

4 years ago

How long will it take a car to go from a complete stop to 44 km/hr if they are accelerating at 5 km/hr?

Aleonysh [2.5K]

Solution :-

Given that ,

Initial Velocity of car = 44km / hr.
Final Velocity of car = 0km / hr.
Acceleration = -5km/hr².

To Find :

Time taken to stop the car .

So , here use first equⁿ of motion which is ,

$\boxed{\red{\bf\dag v = u + at }}$

where ,

v is final Velocity.
u is Initial velocity.
a is acceleration.
t is time taken.

Now , substituting the respective values ,

$\tt:\implies v=u+at$

$\tt:\implies 0 = 44 + (-5)t$

$\tt:\implies -44 = -5t$

$\tt:\implies t=\dfrac{44}{5}$

$\underline{\boxed{\red{\tt \longmapsto Time\:\:=\:\:8.8hrs.}}}$

$\boxed{\green{\bf\pink{\dag} Hence\:time\:taken\:to\:stop\:the\:car\:is\:8.8hrs.}}$

6 0

3 years ago

A 54 kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 1.3 m/s. The acceleration of

Marizza181 [45]

Answer:

$h_{B} = 5.012\, m$

Explanation:

It is assumed that pole vaulter began running at a height of zero. The physical model is formed after the Principle of Energy Conservation:

$K_{A} = K_{B} + U_{B}$

$\frac{1}{2} \cdot m \cdot v_{A}^{2} = \frac{1}{2} \cdot m \cdot v_{B}^{2} + m \cdot g \cdot h_{B}$

The previous expression is simplified and required height is found:

$h_{B} = \frac{1}{2\cdot g} \cdot (v_{A}^{2}-v_{B}^{2})$

$h_{B} = \frac{1}{2 \cdot (9.807\, \frac{m}{s^{2}} )} \cdot [(10\, \frac{m}{s} )^{2}-(1.3\, \frac{m}{s} )^{2}]$

$h_{B} = 5.012\, m$

5 0

4 years ago

frictional force acting on a 0.5 kg object and a floor is 3 N. What is the coefficient of friction between the object and the fl

ikadub [295]

Answer:

μ = 0.6

Explanation:

F = μN

N = mg

F = μmg

3 N = μ*0.5 kg * 9.8 m/s²

μ = 3/(0.5*9.8) = 0.6

7 0

3 years ago

What phases is shown in each of the diagrams above?

taurus [48]

Answer:
A. Interphase
B. Cytokinesis
C. Metaphase
D. Anaphase

Correct me if I was wrong :)

7 0

4 years ago