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vodomira [7]
3 years ago
5

Is there any book for physics.​

Physics
1 answer:
ValentinkaMS [17]3 years ago
3 0

Answer:

which class or course

Explanation:

please tell

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When an object is dropped from the top of a tall cliff, your teacher tells you to use -10 m/s2 for the approximate acceleration
Romashka-Z-Leto [24]

1. Answer: Option C

Displacement is a vector quantity. It has both magnitude as well as direction. Hence, it can be positive,  zero or negative. In the given question, downward direction is taken as negative. This we have come to know from the downward acceleration -10m/s^2. The negative sign indicates that downward direction is considered negative.

Now, since the object is dropped from the top. The displacement should be negative, because the negative represent a downward direction.

2. Answer: velocity

The truck is travelling at 30 mph to the west. From the unit mph (miles per hour), we know its a magnitude of velocity. West indicates the direction. Hence, the given quantity is a vector -velocity.

It is not speed because speed only indicates the magnitude and not the direction. speed is a scalar quantity.

3 0
3 years ago
Read 2 more answers
A) A spaceship passes you at a speed of 0.800c. You measure its length to be 31.2 m .How long would it be when at rest?
rosijanka [135]

Answer:

a

     l_o  =52 \  m

b

      l = 37.13 \ LY

Explanation:

From the question we are told that

    The  speed of the spaceship is  v  =  0.800c

    Here  c is the speed of light with value  c =  3.0*10^{8} \ m/s

    The  length is  l = 31.2 \  m

     The  distance of the star for earth is d = 145 \  light \  years

     The  speed is v_s = 2.90 *10^{8}

     

Generally the from the length contraction equation we have that

       l  =  l_o  \sqrt{1 -[\frac{v}{c } ]}

Now the when at rest the length is  l_o

So  

      l_o =\frac{l}{\sqrt{ 1 - \frac{v^2}{c^2 } } }

      l_o =\frac{ 31.2 }{ \sqrt{1 - \frac{(0.800c ) ^2}{c^2} } }

      l_o=52 \  m

Considering b  

  Applying above equation

            l  =l_o \sqrt{1 -  [\frac{v}{c } ]}

Here l_o  =145 \  LY(light \ years )

So

           l=145 *  \sqrt{1 -  \frac{v_s^2}{c^2 } }

            l =145 *  \sqrt{ 1 - \frac{2.9 *10^{8}}{3.0*10^{8}} }

            l = 37.13 \ LY

4 0
3 years ago
6. State whether each of the following is an example of conduction, convection, and/or radiation. Explain if you think there is
hodyreva [135]
A).  Convection is heating the soup in the pot.
When you stick the spoon into the hot soup,
conduction heats the spoon all the way up to the end.

b).  Water conducts heat a little bit. 
But convection is much more responsible for the
uniform distribution of temperature in the kiddie pool.

c).  The heat from the metal bench conducts directly
to the buttus epidermis when you sit on it.

d).  You feel the heat on your face ... but not on the back of your
neck ... on account of radiation from the fire and the hot grill.
3 0
3 years ago
A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim
Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

F_D = -\frac{1}{2}\rho V^2 C_d A

Where

\rho is the density of the flow

V = Velocity

C_d= Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

F=ma=m\frac{dV}{dt}

Equating both equations we have:

m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A

m(dV)=-\frac{1}{2}\rho C_d A (dt)

\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)

Integrating

\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)

-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t

Here,

V_f = 60mph = 26.82m/s

V_i = 120.7m/s

m= 1600lbf = 725.747Kg

\rho = 1.21 kg/m^3

C_d = 0.3

\Delta t=7s

Replacing:

\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)

-0.029 = -5.4997r^2

r = 2.2963m

d= r*2 = 4.592m \approx 15.065ft

4 0
3 years ago
A person hits a tennis ball with a mass of 0.058 kg against a wall.
horrorfan [7]

Explanation:

Mass of the ball, m = 0.058 kg

Initial speed of the ball, u = 11 m/s

Final speed of the ball, v = -11 m/s (negative as it rebounds)

Time, t = 2.1 s

(a) Let F is the average force exerted on the wall. It is given by :

F=\dfrac{m(v-u)}{t}

F=\dfrac{0.058\times (11-(-11))}{2.1}

F = 0.607 N

(b) Area of wall, A=3\ m^2

Let P is the average pressure on that area. It is given by :

P=\dfrac{F}{A}

P=\dfrac{0.607\ N}{3\ m^2}

P = 0.202 Pa

Hence, this is the required solution.

8 0
3 years ago
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