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kirza4 [7]
3 years ago
5

How does momentum play in tackling?

Physics
1 answer:
shusha [124]3 years ago
5 0

Answer:

When two players are running full speed at each other on a football field they build up their momentum

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klasskru [66]

Answer:

The red car would experience the greatest acceleration.

Explanation:

Newton says that Force equals mass times acceleration or F = ma

We get a = F/m

If we want the greatest acceleration or a, mass or m must be the lowest.

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Which term, taken from the celestial sphere, gives its name to a.m. and p.m.?
DaniilM [7]

Options are. Zenith, Great circle, Equinox, or Meridan
3 0
3 years ago
Which weather condition commonly occurs along a cold front
aleksklad [387]
Precipitation is your answer. according to my science textbook, in a cold front, warm air and cold air interact and cause precipitation.
7 0
3 years ago
Read 2 more answers
A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It s
Zarrin [17]

Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± \frac{1}{2} at^2               ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± \frac{1}{2} at^2               ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

<em>From the question;</em>

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s   (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - \frac{1}{2} at^2

h = 26.54(1.7) - \frac{1}{2} (10)(1.7)^2

h = 45.118 - 14.45

h = 30.67m

Therefore, the vertical distance is 30.67m        

7 0
3 years ago
A department store sells an ""astronomical telescope"" with an objective lens of 30 cm focal length and an eyepiece lens of foca
Yuliya22 [10]

Answer:

The magnifying power of this telescope is (-60).

Explanation:

Given that,

The focal length of the objective lens of an astronomical telescope, f_o=30\ cm

The focal length of the eyepiece lens of an astronomical telescope, f_e=5\ mm=0.5\ cm

To find,

The magnifying power of this telescope.

Solution,

The ratio of focal length of the objective lens to the focal length of the eyepiece lens is called magnifying of the lens. It is given by :

m=\dfrac{-f_o}{f_e}

m=\dfrac{-30}{0.5}

m = -60

So, the magnifying power of this telescope is 60. Therefore, this is the required solution.

7 0
3 years ago
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