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Eva8 [605]
3 years ago
12

In a Little League baseball game, the 145 g ball reaches the batter with a speed of 15.0 m/s. The batter hits the ball, and it l

eaves his bat with a speed of 20.0 m/s in exactly the opposite direction. What is the magnitude of the impulse delivered by the bat to the ball
Physics
1 answer:
Oduvanchick [21]3 years ago
3 0

Answer: 5.075Ns

Explanation:

Given the following :

Mass of ball = 145g

Initial Speed of ball = 15m/s

Final speed of ball when hit by the batter = - 20m/s ( Opposite direction)

The impulse of a body is represented using the relation:

Force(f) * time(t) = mass (m) * (final Velocity(V) - initial velocity(u))

Therefore, using:

m(v - u) = impulse

Mass of ball = 145 / 1000 = 0.145kg

Impulse = 0.145(- 20 - 15)

Impulse = 0.145(-35)

Impulse = 5.075Ns

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A Styrofoam box has a surface area of 0.73 m and a wall thickness of 2.0 cm. The temperature of the inner surface is 5.0°C, and
Elis [28]

Answer:

K = 0.076 W/m K

Explanation:

Heat required to melt the complete ice is given as

Q = mL

here we have

m = 5.0 kg

L = 3.35 \times 10^5 J/kg

now we have

Q = (5 kg)(3.35 \times 10^5)

Q = 1.67 \times 10^6 J

now the power required to melt ice in 8.4 hours is

P = \frac{Q}{t} = \frac{1.67\times 10^6}{8.4 \times 3600 s}

P = 55.4 Watt

now by formula of conduction we know

P = \frac{KA(\Delta T)}{x}

now we have

55.4 = \frac{K(0.73 m^2)(25 - 5)}{0.02}

K = 0.076 W/m K

5 0
3 years ago
a toy car is wound up and released on the floor. it accelerates at a rate of 0.4 m/s/s . the mass of the care is 3kg. what is th
EastWind [94]

Answer:

1.2N

Explanation:

f=m×a

m=3kg

a=.4m/s/s

3×.4=1.2

kg×m/s/s = N

8 0
3 years ago
At T = 0C and p = 1000mb, 1 g of dry air receives an amount of heat during an isochoric process. It is then observed that its p
Rufina [12.5K]

Answer:

ΔT = 13.65° C

ΔQ = 13.7 J

Explanation:

First we will find the final temperature of air by using equation of state:

P₁V₁/T₁ = P₂V₂/T₂

For Isochoric Process, V₁ = V₂

Therefore,

P₁/T₁ = P₂/T₂

T₂ = P₂T₁/P₁

where,

T₂ = Final Temperature = ?

P₂ = Final Pressure = 1050 mb

P₁ = Initial Temperature = 1000 mb

T₁ = Initial Temperature = 0°C = 273 k

Therefore,

T₂ = (1050 mb)(273 K)/(1000 mb)

T₂ = 286.65 K

Change in Temperature = ΔT = T₂ - T₁

ΔT = 286.65 K - 273 K

<u>ΔT = 13.65° C</u>

<u></u>

The first law of thermodynamics can be written as:

ΔQ = ΔU + W

where,

ΔQ = heat absorbed

ΔU = change in internal energy = mCΔT

W = Work Done = 0 (in case of isochoric process)

Therefore.

ΔQ = mCΔT

where,

m = mass of air = 1 g = 1 x 10⁻³ kg

C = specific heat of dry air = 1003.5 J/kg.°C

Therefore,

ΔQ = (1 x 10⁻³ kg)(1003.5 J/kg.°C)(13.65°C)

<u>ΔQ = 13.7 J</u>

7 0
3 years ago
The measure of the force with which air molecules push on a surface is called
alex41 [277]
It is called air pressure
7 0
3 years ago
Particle A of charge 2.76 10-4 C is at the origin, particle B of charge -6.54 10-4 C is at (4.00 m, 0), and particle C of charge
Vanyuwa [196]

Answer:

a) F_net = 30.47 N ,   θ = 10.6º

b)  Fₓ = 29.95 N

Explanation:

For this exercise we use coulomb's law

          F₁₂ = k k \frac{ q_{1}  \  q_{2} }{ r^{2} }

the direction of the force is on the line between the two charges and the sense is repulsive if the charges are equal and attractive if the charges are different.

As we have several charges, the easiest way to solve the problem is to add the components of the force in each axis, see attached for a diagram of the forces

X axis

        Fₓ = F_{bc x}

Y axis  

       F_{y}Fy = F_{ab} - F_{bc y}

let's find the magnitude of each force

     F_{ab} = 9 10⁹ 2.76 10⁻⁴ 1.02 10⁻⁴ / 3²

      F_{ab} = 2.82 10¹ N

      F_{ab} = 28.2 N

   

      F_{bc} = 9 10⁹ 6.54 10⁻⁴ 1.02 10⁻⁴ / 4²

      F_{bc} = 3.75 10¹  N

       F_{bc} = 37.5 N

let's use trigonometry to decompose this force

      tan θ = y / x

      θ = tan⁻¹ and x

       θ= tan⁻¹ ¾

      θ = 37º

let's break down the force

      sin 37 = F_{bcy} / F_{bc}

      F_{bcy} = F_{bc} sin 37

      F_{bcy} = 37.5 sin 37

      F_{bcy} = 22.57 N

      cos 37 = F_{bcx} /F_{bc}

      F_{bcx} = F_{bc} cos 37

      F_{bcx} = 37.5 cos 37

      F_{bcx} = 29.95 N

let's do the sum to find the net force

X axis

        Fₓ = 29.95 N

Axis y

        Fy = 28.2 -22.57

        Fy = 5.63 N

we can give the result in two ways

a)  F_net = Fₓ i ^ + F_{y} j ^

    F_net = 29.95 i ^ + 5.63 j ^

b) in the form of module and angle

let's use the Pythagorean theorem

    F_net = \sqrt{ F_{x}^2 + F_{y}^2 }

    F_net = √(29.95² + 5.63²)

     F_net = 30.47 N

we use trigonometry for the direction

      tan θ= \frac{ F_{y}  }{  F_{x} }

       

      θ = tan⁻¹ \frac{ F_{y}  }{  F_{x} }

      θ = tan⁻¹ (5.63 / 29.95)

      θ = 10.6º

3 0
3 years ago
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