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Pavel [41]
3 years ago
11

The vertical component of the acceleration of a projectile remains constant during the entire trajectory of the projectile.

Physics
1 answer:
WARRIOR [948]3 years ago
6 0

Answer: True

Explanation:

I had this question

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I NEED HELP SOLVING THIS!!!!!!!!!!!!
Neko [114]

Answer:

Yes

Explanation:

The given parameters are;

The speed with which the fastball is hit, u = 49.1 m/s (109.9 mph)

The angle in which the fastball is hit, θ = 22°

The distance of the field = 96 m (315 ft)

The range of the projectile motion of the fastball is given by the following formula

Range = \dfrac{u^2 \times sin(2\cdot \theta)}{g}

Where;

g = The acceleration due to gravity = 9.81 m/s², we have;

Range = \dfrac{49.1^2 \times sin(2\times22^{\circ})}{9.81} \approx 170.71 \ m

Yes, given that the ball's range is larger than the extent of the field, the batter is able to safely reach home.

7 0
3 years ago
How can a calculated height be greater than an actual height?
sammy [17]

Answer:

mesuring heigh and weight is important

3 0
3 years ago
What do biologist geologist have in common how are they different
Delicious77 [7]
The prefix for Bio is life so a biologist studies life of living organisms
The prefix for Geo is earth so it would be the study of earth.
What is in common would be animals and plants
what is different is an Geologist studies only earth Biologist studies everything living. I hope this was helpful and I didn't confuse you.

6 0
3 years ago
Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac
Serjik [45]
The strength of the gravitational field is given by:
g= \frac{GM}{r^2}
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m

And now we can use the previous equation to calculate the field strength at that altitude:
g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2}  = 8.95 m/s^2

And we can see this value is a bit less than the gravitational strength at the surface, which is g_s = 9.81 m/s^2.
4 0
3 years ago
What is the wavelength of a 6.00*10^2 Hz sound wave in air at 20C
Alex Ar [27]

Answer:

Solution

λ=v/n

Here, v=344 m s−1

n=22 MHz =22×106 Hz

λ=344/22×106=15.64×10−6m=15.64μm.

5 0
2 years ago
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