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LenaWriter [7]
3 years ago
11

6. A football is kicked with an initial speed of 10.2 m/s at an angle of 40.00 above the horizontal. It lands on the ground 2.12

s later. Determine the velocity of the football at the pinnacle of its trajectory.
Physics
1 answer:
igor_vitrenko [27]3 years ago
7 0

Answer:

7.81 m/s

Explanation:

Given,

initial speed, u = 10.2 m/s

angle of inclination, θ = 40°

time, t = 2.12 s

Horizontal component of the velocity:

u_x = u cos \theta

u_x = 10.2\times cos 40^0

u_x = 7.81 m/s

In projectile motion horizontal component of the velocity remain same at every point because there is no acceleration.

So, Velocity at the Pinnacle is equal to 7.81 m/s

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10,500 J of GPE, weight 539 N, how tall is the hill you are sitting on?
maks197457 [2]

Answer:

19.48 m

Explanation:

Gravitational potential energy = mgh

Current weight = 539 N

Weight = mg = 539 N

Mass x Acceleration = 539 N

Mass x 9.81 = 539

Mass = 54.94 kg

Gravitational potential energy = mgh = 10500 J

                    54.94 x 9.81 x h = 10500

                               h = 19.48 m

Height of sitting = 19.48 m

6 0
3 years ago
What did solomon asch discover in his famous experiment on judging the lengths of lines?
BaLLatris [955]

Answer:

Asch (1956) found that group size influenced whether subjects conformed. The bigger the majority group (no of confederates), the more people conformed, but only up to a certain point.

Explanation:

4 0
3 years ago
An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass
gizmo_the_mogwai [7]

Answer:

Explanation:

First of all we shall find the velocity at equilibrium point of mass 1.2 kg .

It will be ω A , where ω is angular frequency and A is amplitude .

ω = √ ( k / m )

= √ (170 / 1.2 )

= 11.90 rad /s

amplitude A = .045 m

velocity at middle point ( maximum velocity ) = 11.9 x .045 m /s

= .5355 m /s

At middle point , no force acts so we can apply law of conservation of momentum

m₁ v₁ = ( m₁ + m₂ ) v

1.2 x .5355 = ( 1.2 + .48 ) x v

v = .3825 m /s

= 38.25 cm /s

Let new amplitude be A₁ .

1/2 m v² = 1/2 k A₁²

( 1.2 + .48 ) x v² = 170 x A₁²

( 1.2 + .48 ) x .3825² = 170 x A₁²

A₁ = .0379 m

New amplitude is .0379 m

7 0
3 years ago
In the United States, where can volcanoes be found?
Sedbober [7]
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7 0
3 years ago
A large solar panel on a spacecraft in Earth orbit produces 1.0 kW of power when the panel is turned toward the sun. What power
Mandarinka [93]

Answer:

e*P_s = 11 W

Explanation:

Given:

- e*P = 1.0 KW

- r_s = 9.5*r_e

- e is the efficiency of the panels

Find:

What power would the solar cell produce if the spacecraft were in orbit around Saturn

Solution:

- We use the relation between the intensity I and distance of light:

                                  I_1 / I_2 = ( r_2 / r_1 ) ^2

- The intensity of sun light at Saturn's orbit can be expressed as:

                                  I_s = I_e * ( r_e / r_s ) ^2

                                  I_s = ( 1.0 KW / e*a) * ( 1 / 9.5 )^2

                                  I_s = 11 W / e*a

- We know that P = I*a, hence we have:

                                  P_s = I_s*a

                                  P_s = 11 W / e

Hence,                       e*P_s = 11 W

3 0
3 years ago
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