All categories

3 years ago

A football is kicked from a tee at 12 m/s at 72° above the horizontal. What is the maximum height of the football

Physics

1 answer:

ale4655 [162]3 years ago

6 0

The maximum height is reached when the vertical component of the velocity is zero.

vertical direction:
acceleration: a = -g = -9.81m/s²
velocity: v = -g*t + v₀
position: y = -0.5*g*t² + v₀*t + y₀

For v= 0:
0 = -g*t + v₀ => t = v₀/g
Insert into position equation gives:
y(max) = (-0.5*v₀²/g) + (v₀²/g) + y₀ = (0.5*v₀²/g) + y₀

You might be interested in

Choose the statement(s) that is/are true about the ratio \frac{C_p}{C_v} C p C v for a gas? (Ii) This ratio is the same for all

Blababa [14]

Answer:

(i) false

(ii) true

(iii) true

(iv) false

Explanation:

(i) The ratio of Cp and Cv is not constant for all the gases. It is because the value of cp and Cv is different for monoatomic, diatomic and polyatomic gases.

So, this is false.

(ii) For monoatomic gas

Cp = 5R/2, Cv = 3R/2

So, thier ratio

Cp / Cv = 5 / 3 = 1.67

This statement is true.

(iii) for diatomic gases

Cp = 7R/2, Cv = 5R/2

Cp / Cv = 7 / 5 = 1.4

This statement is true.

(iv) It is false.

6 0

3 years ago

You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has

vovikov84 [41]

Answer:8.75 s,

136.89 m

Explanation:

Given

Initial velocity $=70 mph\approx 31.29 m/s$

velocity after 5 s is $30 mph\approx 13.41 m/s$

Therefore acceleration during these 5 s

$a=\frac{v-u}{t}$

$a=\frac{13.41-31.29}{5}=-3.576 m/s^2$

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

$0=31.29-3.576\times t$

$t=\frac{31.29}{3.576}=8.75 s$

(b)total distance traveled before stoppage

$v^2-u^2=2as$

$0^2-31.29^2=2\times (-3.576)\cdot s$

s=136.89 m

3 0

3 years ago

Explain what a concentration gradient is and what it means for a molecule to diffuse down its

MArishka [77]

Answer:

Diffusing the gradient ensures that most of the molecules in high concentration zone will wind up in the previously low concentration by the spontaneous movement of small molecules.

Explanation:

A gradient of concentration is the difference between in concentration of one place / area substance to different area. Having a molecule flow down its concentration gradient means moving the molecules from hypotonic areas to the concentration hypertonic areas

Diffusing the gradient ensures that most of the molecules in high concentration zone will wind up in the previously low concentration by the spontaneous movement of small molecules.

7 0

3 years ago

a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its

lilavasa [31]

Answer:

D/H =15

Explanation:

We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
At this point, we can find the value of H, applying the following kinematic equation:

$v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)$

If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

$H = \frac{v_{0} ^{2} }{2*g} (2)$

We can use the same equation, to find the value of D, as follows:

$v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)$

In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
When we replace these values in (1), we find that v₁ = -v₀.
Replacing in (3), we have:

$(4*v_{0})^{2} - (-v_{0}) ^{2} = 2* g* D\\ \\ 15*v_{0}^{2} = 2*g*D$

Solving for D:

$D = \frac{15*v_{0} ^{2} }{2*g}$

From (2) we know that H can be expressed as follows:

$H = \frac{v_{0} ^{2} }{2*g}$

⇒ D = 15 * H

$\frac{D}{H} = 15$

3 0

2 years ago

Define orbital velocity

sveta [45]

the Orbital Velocity is the velocity sufficient to cause a natural or artificial satellite to remain in orbit. Inertia of the moving body tends to make it move on in a straight line, while gravitational force tends to pull it down. The orbital path, elliptical or circular, representing a balance between gravity and inertia, and it follows a rue that states that the more massive the body at the centre of attraction is, the higher is the orbital velocity for a particular altitude or distance.

8 0

3 years ago