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masya89 [10]
2 years ago
12

During a baseball game, a batter hits a high pop-up. if the ball remains in the air for a total of 6.0 s, how high does it rise?

( Assume Accel. = 10 m/s^2 )
Physics
1 answer:
andrew11 [14]2 years ago
5 0
  • Time=6s
  • Vertex height's time= 6/2=3s
  • Acceleration due to gravity=10m/s^2

Apply second equation of kinematics

\\ \rm\hookrightarrow s=ut+\dfrac{1}{2}gt^2

\\ \rm\hookrightarrow s=1/2(10)(3)^2

\\ \rm\hookrightarrow s=5(9)

\\ \rm\hookrightarrow s=45m

It rises 45m

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Answer:

18

Explanation:

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3 years ago
Technician A says that other temperature sensors that operate like the ECT include transmission fluid temperature (TFT), and eng
PIT_PIT [208]

Answer Choices:

a. Technician A only

b. Technician B only

c. Both technicians A and B

d. Neither technician A nor B

Explanation:

a. Technician A only

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3 years ago
What force is required to accelerate a 1840 kg car from 4.77 m/s to 23.5 m/s,
neonofarm [45]
A :-) for this question , we should apply
a = v - u by t
Given - u = 4.77 m/s
v = 23.5 m/s
t = 5.18 m/s
Solution -
a = v - u by t
a = 23.5 - 4.77
a = 28.27 m/s^2

.:. The acceleration is 28.27 m/s^2
7 0
3 years ago
The Assignment: A fixed quantity of an ideal gas (R 0.28 kJ/kgK; Cv-0.71kJ/kgK) is expanded from an initial condition of 35 bar,
Nikolay [14]

Answer:

Index of expansion: 4.93

Δu = -340.8 kJ/kg

q = 232.2 kJ/kg

Explanation:

The index of expansion is the relationship of pressures:

pi/pf

The ideal gas equation:

p1*v1/T1 = p2*v2/T2

p2 = p1*v1*T2/(T2*v2)

500 C = 773 K

20 C = 293 K

p2 = 35*0.1*773/(293*1.3) = 7.1 bar

The index of expansion then is 35/7.1 = 4.93

The variation of specific internal energy is:

Δu = Cv * Δt

Δu = 0.71 * (20 - 500) = -340.8 kJ/kg

The first law of thermodynamics

q = l + Δu

The work will be the expansion work

l = p2*v2 - p1*v1

35 bar = 3500000 Pa

7.1 bar = 710000 Pa

q = p2*v2 - p1*v1 + Δu

q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg

7 0
3 years ago
In a physics lab, Asha is given a 10.7 kg uniform rectangular plate with edge lengths 67.3 cm by 53.5 cm . Her lab instructor re
Leya [2.2K]

Answer:

I=2.6363\ kg.m^2

Explanation:

Given:

dimension of uniform plate, (0.673\times 0.535)\ m^2

mass of plate, m=10.7\ kg

Now we find the moment of inertia about the center of mass of the rectangular plate is given as:

I_{cm}=\frac{1}{12} \times m(L^2+B^2)

where:

L= length of the plate

B= breadth of the plate

I_{cm}=\frac{1}{12} \times 10.7\times(0.673^2+0.535^2)

I_{cm}=0.6591\ kg.m^2

We know that the center of mass of the rectangular plane is at its geometric center which is parallel to the desired axis XX' .

Now we find the distance between the center of mass and the corner:

s=\frac{\sqrt{ (0.673^2+0.535^2)}}{2}

s=0.4299\ m

Now using parallel axis theorem:

I=I_{cm}+m.s^2

I=0.6591+10.7\times 0.4299^2

I=2.6363\ kg.m^2

6 0
3 years ago
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