PHYSICS COLLEGE

A baseball leaves the bat with a speed of 40 m/s at an angle of 35 degrees. A 12m tall fence is placed 130 m from the point the ball was struck. Assuming the batter hit the ball 1m above ground level, does the ball go over the fence? If not, how does the ball hit the fence? If yes, how far beyond the fence does the ball land?

Answers

Answer 1

Answer:

Answer:

The ball land at 3.00 m.

Explanation:

Given that,

Speed = 40 m/s

Angle = 35°

Height h = 1 m

Height of fence h'= 12 m

We need to calculate the horizontal velocity

Using formula of horizontal velocity

$V_(x)=V_(i)\cos\theta$

$V_(x)=40*\cos35$

$V_(x)=32.76\ m/s$

We need to calculate the time

Using formula of time

$t = (d)/(v)$

$t=(130)/(32.76)$

$t=3.96\ sec$

We need to calculate the vertical velocity

$v_(y)=v_(y)\sin\theta$

$v_(y)=40*\sin35$

$v_(y)=22.94\ m/s$

We need to calculate the vertical position

Using formula of distance

$y(t)=y_(0)+V_(i)t+(1)/(2)gt^2$

Put the value into the formula

$y(3.96)=1+22.94*3.96+(1)/(2)*(-9.8)*(3.96)^2$

$y(3.96)=15.00\ m$

We need to calculate the distance

$s = y-h'$

$s=15.00-12$

$s=3.00\ m$

Hence, The ball land at 3.00 m.

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Two bodies, one hot and the other cold kept in vacuum.what will happen to the tempreture of bodies after some time.

Answers

Hot body will lose heat from it, and that heat will goes out from it through radiation, so it's temperature will decrease after some time.

In same manner, cold body will take the heat, and it's temperature will increase

Hope this helps!

A Venturi tube may be used as the inlet to an automobile carburetor. If an inlet pipe with a diameter of 2.0 cm diameter narrows to diameter of 1.0 cm, determine the pressure drop in the constricted section for an initial airflow of 3.0 m/s in the 2-cm section? (Assume air density is 1.25 kg/m

Answers

The pressure drop is equal to 80.99 Pa

Given information:

d1 = 2 cm = 0.02 m

d2 = 1 cm = 0.01 m

v = 3 m/s

p = 1.25 kg/m^3

Here we use Bernoulli's principle for the Venturi Tube:

Calculation of pressure drop:

$P1 - P2 = ((v^2* p)/ 2)* ((A1^2/ A2^2)-1)\n\nP1 - P2 = \Delta P = ((v1^2* p)/ 2)* ((A1^2/ A2^2)-1)$

Now the following formula for area calculation should be used:

$A1 = (\pi* d1^2)/ 4 = (\pi* (0.02 m)^2)/ 4 = 0.00031 m^2\n\nA2 = (\pi* (0.01 m)^2)/ 4 = 0.000079 m^2\n\n\Delta P = ((3 m/s)^2 *1.25 kg/m^3)/ 2) * ((0.00031 m^2)^2/(0.000079 m^2)^2)-1)$

= 80.99

Find out more information about the Pressure here: brainly.com/question/356585?referrer=searchResults

Answer:

the pressure drop is equal to 80.99 Pa

Explanation:

we have the following data:

d1 = 2 cm = 0.02 m

d2 = 1 cm = 0.01 m

v = 3 m/s

p = 1.25 kg/m^3

ΔP = ?

For the calculation of the pressure drop we will use Bernoulli's principle for the Venturi Tube:

P1 - P2 = ((v^2*p)/2)*((A1^2/A2^2)-1)

where A = area

P1 - P2 = ΔP = ((v1^2*p)/2)*((A1^2/A2^2)-1)

for the calculation of the areas we will use the following formula:

A1 = (pi*d1^2)/4 = (pi*(0.02 m)^2)/4 = 0.00031 m^2

A2 = (pi*(0.01 m)^2)/4 = 0.000079 m^2

ΔP = ((3 m/s)^2*1.25 kg/m^3)/2)*((0.00031 m^2)^2/(0.000079 m^2)^2)-1) = 80.99 N/m^2 = Pa

True Or False for each question

Answers

7 true

8 false

9 false

10 false

11 false

12 true

13 true

hope this helps!

Callisto, one of Jupiter's moons, has an orbital period of 16.69 days and its distance from Jupiter is 1.88*10^6 km. What is Jupiter's mass?

Answers

Answer:

The Jupiter´s mass is approximately 1.89*10²⁷ kg.

Explanation:

The only force acting on Calisto while is rotating around Jupiter, is the gravitational force, as defined by the Newton´s Universal Law of Gravitation:

Fg = G*mc*mj / rcj²

where G = 6.67*10⁻¹¹ N*m²/kg², mc= Callisto´s mass, mj= Jupiter´s mass, and rcj = distance from Jupiter for Callisto= 1.88*10⁹ m.

At the same time, there exists a force that keeps Callisto in orbit, which is the centripetal force, that actually is the same gravitational force we have already mentioned.

This centripetal force is related with the period of the orbit, as follows:

Fc = mc*(2*π/T)²*rcj.

In order to be consistent in terms of units, we need to convert the orbital period, from days to seconds, as follows:

T = 16.69 days* 86,400 (sec/day) = 1.44*10⁶ sec.

We have already said that Fg= Fc, so we can write the following equality:

G*mc*mj / rcj² = mc*(2*π/T)²*rcj

Simplifying common terms, and solving for mj, we get:

mj = 4*π²*(1.88*10⁹)³m³ / ((1.44*10⁶)² m²*6.67*10⁻11 N*m²/kg²)

mj = 1.89*10²⁷ kg.

Answer: Mass of Jupiter ~= 1.89 × 10^23 kg

Explanation:

Given:

Period P= 16.69days × 86400s/day= 1442016s

Radius of orbit a = 1.88×10^6km × 1000m/km

r = 1.88 × 10^9 m

Gravitational constant G= 6.67×10^-11 m^3 kg^-1 s^-2

Applying Kepler's third law, which is stated mathematically as;

P^2 = (4π^2a^3)/G(M1+M2) .....1

Where M1 and M2 are the radius of Jupiter and callisto respectively.

Since M1 >> M1

M1+M2 ~= M1

Equation 1 becomes;

P^2 = (4π^2a^3)/G(M1)

M1 = (4π^2a^3)/GP^2 .....3

Substituting the values into equation 3 above

M1 = (4 × π^2 × (1.88 × 10^9)^3)/(6.67×10^-11 × 1442016^2)

M1 = 1.89 × 10^27 kg

A neutral metal ball is suspended by a string. A positively charged insulating rod is placed near the ball, which is observed to be attracted to the rod. This is because:____________. a. the ball becomes negatively charged by induction
b. the ball becomes positively charged by induction
c. the string is not a perfect insulator
d. there is a rearrangement of the electrons in the ball
e. the number of electrons in the ball is more than the number in the rod

Answers

Answer:

d. there is a rearrangement of the electrons in the ball

Explanation:

Inside the neutral metal ball, there are equal no. of positive charges (protons) and negative charges (electrons). Normally, the charges are distributed evenly throughout the ball.

However, when the positively charged insulating rod is brought near, since positive charges and negative charges attract each other, the electrons (-ve charges) in the metal ball moves towards the side nearest to the rod. The metal ball gets attracted to the rod.

a and b are not correct because the rod is insulating, so electrons cannot be transferred between them to induce a net charge in the metal ball. the no. of electrons is unrelated to the attraction between opposite charges , so e is incorrect as well.

A glass plate 2.95 mmmm thick, with an index of refraction of 1.60, is placed between a point source of light with wavelength 600 nmnm (in vacuum) and a screen. The distance from source to screen is 1.25 cm. How many wavelengths are there between the source and the screen?