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Ber [7]
2 years ago
12

A baseball is launched horizontally from a height of 1.8 m. The baseball travels 0.5 m before hitting the ground. How fast is th

e baseball moving horizontally, rounded to
Physics
1 answer:
zysi [14]2 years ago
4 0

Answer:

0.83 m/s

Explanation:

FIrst of all, we have to find the time of flight, i.e. the time the baseball needs to reach the ground. This can be done by using the equation for the vertical motion:

h=ut+\frac{1}{2}gt^2

where

h is the initial height

u = 0 is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Substituting h = 1.8 m and solving for t,

t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(1.8)}{9.8}}=0.61 s

We know that the horizontal distance travelled by the ball is

d = 0.5 m

Therefore, we can find the horizontal velocity (which is constant during the whole motion):

v= \frac{d}{t}=\frac{0.5}{0.60}=0.83 m/s

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Can someone please help with this. "A ball is thrown vertically upward with a speed of 26.6 m/s. How high does it rise? The acce
lubasha [3.4K]

v^2=u^2+2as

0=26.6^2-2x9.8xs

-26.6^2/-2x9.8=s

calculator

8 0
3 years ago
The core of a 400 Hz aircraft transformer has a net cross-sectional area of 13 cm2. The maximum flux density is 0.9 T, and there
jenyasd209 [6]

Answer:

32.76 Volt

Explanation:

frequency, f = 400 Hz

Area of crossection, A = 13 cm²

Maximum flux density, B = 0.9 tesla

Number of turns in secondary coil, N = 70

Let the maximum induced voltage is e.

According to the Faraday's law of electromagnetic induction, the induced emf is equal to the rate of change of magnetic flux.

e = dФ/dt

e=\frac{NBA}{t}

Time is defined as the reciprocal of frequency.

So, e = N B A f

e = 70 x 0.9 x 13 x 10^-4 x 400

e = 32.76 volt

4 0
2 years ago
If a car's engine gives off 65% thermal energy, what is the maximum efficiency?
stich3 [128]

Answer:

35%

Explanation:

The car's engine gives off 65% thermal energy

So only 35 % is converted into mechanical energy .

input heat = Q₁ = 100

output heat = Q₂ = 65

Work output = Q₁ - Q₂ = W

W = 100 - 65 = 35

Efficiency = W / Q₁ X 100

= (35/ 100) X 100

= 35%.

5 0
2 years ago
When exercising in the heat, which of the following hydration strategies is best for temperature regulation during an event (e.g
Olegator [25]

Answer:

Your question was incomplete so here is the complete question and answer.

Q. When exercising in the heat, which of the following hydration strategies is best for temperature regulation during an event (e.g., 10K race)

a) plain water

b) 5-7 percent glucose solution

c) Glucose polymer solution of 6-8 percent

d) There appears to be no difference among these different forms of hydration techniques relative to temperature regulation.

Ans. d) There appears to be no difference among these different forms of hydration techniques relative to temperature regulation.

Explanation:

Temperature Regulation is an important phenomenon for the person exposed to extreme hot conditions or weather. Exercising in hot conditions increase the body temperature. Greater and intense exercise, greater the production of heat. Then the heat dissipation takes place in the form of excessive sweating which results in dehydration. That was just the brief overview of temperature regulation. Above mentioned techniques are equally good hydration techniques so there is no difference at all. You can have a plain water or glucose solutions of above mentioned percentages.

3 0
3 years ago
The component of the external magnetic field along the central axis of a 78-turn circular coil of radius 34.0 cm decreases from
grigory [225]

Answer:

Induced current, I = 18.88 A

Explanation:

It is given that,

Number of turns, N = 78

Radius of the circular coil, r = 34 cm = 0.34 m

Magnetic field changes from 2.4 T to 0.4 T in 2 s.

Resistance of the coil, R = 1.5 ohms

We need to find the magnitude of the induced current in the coil. The induced emf is given by :

\epsilon=-N\dfrac{d\phi}{dt}

Where

\dfrac{d\phi}{dt} is the rate of change of magnetic flux,

And \phi=BA

\epsilon=-NA\dfrac{dB}{dt}

\epsilon=-78\times \pi (0.34)^2\dfrac{(0.4-2.4)}{2}

\epsilon=28.32\ V

Using Ohm's law, \epsilon=I\times R

Induced current, I=\dfrac{\epsilon}{R}

I=\dfrac{28.32}{1.5}

I = 18.88 A

So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.

5 0
2 years ago
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