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Rudik
14 days ago
6

The crossbar of a goal post on a professional American football field sits at a height of 3.05 meters (10 feet) above the field.

A football is placed on the ground at midfield so that it is 54.9 meters (60 yards) horizontally from the position of the crossbar. The ball is then kicked so that its initial velocity is 26.0 m/s (58.2 mph) at an angle of 35 degrees above the horizontal. Neglecting air friction, does the ball make it over the goal post and by how much?
A. Yes, the ball passes 2.83 meters above the crossbar
B. No, the ball passes 1.57 meters below the crossbar
C. No, the ball hits the ground before it reaches the crossbar
D No, the ball hits the crossbar
E. Yes, the ball passes 8.22 meters above the crossbar
Physics
1 answer:
Sav [2.2K]14 days ago
5 0

Answer:

A. Yes, the ball clears the crossbar by 2.83 meters

Explanation:

This situation pertains to projectile motion.

The horizontal velocity component of the ball is calculated as 26 cos 35 = 21.3 m/s

The vertical velocity component is 26 sin 35 = 14.9 m/s

The time taken to travel the horizontal distance to the goalpost, which is 54.9 m, is:

= distance / horizontal speed

= 54.9 / 21.3

= 2.577 seconds.

The vertical distance achieved during this time is:

h = ut - 1/2 gt², where u is the initial vertical velocity, and t = 2.577 seconds.

h = 14.9 x 2.577 - 0.5 x 9.8 x (2.577)²

= 38.39 - 32.54

= 5.85 m

Thus, the ball surpasses the crossbar by 5.85 - 3.05 = 2.8 m

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a fixed mass of a n ideal gas is heated from 50 to 80C at a constant pressure at 1 atm and again at a constant pressure of 3 atm
inna [2210]

Answer:

The required energy remains identical in both scenarios since the specific heat capacity (Cp) does not change with varying pressure.

Explanation:

Given;

initial temperature, t₁ = 50 °C

final temperature, t₂ = 80 °C

Temperature change, ΔT = 80 °C - 50 °C = 30 °C

Pressure for scenario one = 1 atm

Pressure for scenario two = 3 atm

The energy needed in both scenarios is expressed as;

Q = M*C_p*\delta T

Where;

Cp denotes specific heat capacity, which only varies with temperature and remains unaffected by pressure.

Hence, the energy required remains the same for both scenarios since specific heat capacity (Cp) is pressure-independent.

8 0
1 month ago
A spring is stretched 6 in by a mass that weighs 8 lb. The mass is attached to a dashpot mechanism that has a damping constant o
Ostrovityanka [2208]

Response:

y= 240/901 cos 2t+ 8/901 sin 2t

Clarification:

To determine mass m=weight/g

  m=8/32=0.25

To calculate the spring constant

Kx=mg    (with c=6 inches and mg=8 pounds)

K(0.5)=8               (6 inches converts to 0.5 feet)

K=16 lb/ft

The governing equation for the spring-mass system is

my''+Cy'+Ky=F  

Inserting the known values yields

0.25 y"+0.25 y'+16 y=4 cos 20 t  ----(1) (given C=0.25 lb.s/ft)

Assuming the steady state equation for y is

y=A cos 2t+ B sin 2t

To determine constants A and B, we must equate this with equation 1.

Next, we find y' and y" by differentiating with respect to t.

y'= -2A sin 2t+2B cos 2t

y"=-4A cos 2t-4B sin 2t

Now, substitute the values of y", y' and y into equation 1

0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t

By comparing coefficients on both sides

30 A+ B=8

A-30 B=0

From this, we find

A=240/901 and B=8/901

Thus, the steady state response

y= 240/901 cos 2t+ 8/901 sin 2t

6 0
10 days ago
In order for the ball to be able to make a complete circle around the peg, there must be sufficient speed at the top of its arc
kicyunya [2264]

Answer:

Explanation:

Let T represent the tension in the swing.

At the peak mg-T=\frac{mv^2}{r}

where v denotes the velocity needed to maintain the circular motion.

r equals the distance from the rotation point to the center of the ball, which is L+\frac{d}{2} (with d being the ball's diameter).

The threshold velocity can be expressed as mg-0=\frac{mv^2}{r}

To determine the velocity at the bottom, we can use energy conservation principles at both the top and bottom positions.

At the top E_T=mg\times 2L+\frac{mv^2}{2}

Energy at the bottom E_b=\frac{mv_0^2}{2}

By comparing the two states using conservation of energy, we find v_0^2=4gr+gr

v_0^2=5gr

v_0=\sqrt{5gr}

v_0=\sqrt{5g\left ( \frac{d}{2}+L\right )}

6 0
16 days ago
Can pockets of vacuum persist in an ideal gas? Assume that a room is filled with air at 20∘C and that somehow a small spherical
kicyunya [2264]

Answer:

The time required is 20 μs

Explanation:

Here is the data provided:

temperature = 20°C  = 293 K

radius = 1 cm

atomic mass of air = 29 u

To determine

the duration for air to refill the vacuum space

solution:

We calculate the root mean square velocity of air particles. This can be expressed as:

\frac{1}{2}mv^2 = \frac{3}{2}RT

where m indicates mass, t is temperature, v is speed, and R is the ideal gas constant, which is approximately 8.3145 (kg·m²/s²)/K·mol.

v = \sqrt{\frac{3RT}{M} }............................1

v = \sqrt{\frac{3(8.314)293}{29*10{-3}kg} }

Resulting in v = 501.99 m/s.

<pNow, to cover the distance of 1 cm,<pThe duration needed for air is calculated as:

time taken = \frac{r}{v}

which gives us:

time taken = \frac{1*10^{-2}m}{501.99}

so, time taken = 19.92 × 10^{6} seconds = 20μs.

Thus, the required time is 20 μs.

3 0
1 month ago
A 1000-kg car is moving along a straight road down a 30∘30∘ slope at a constant speed of 20.0m/s20.0m/s. What is the net force a
Softa [2035]

The overall force acting on the vehicle is zero

Explanation:

Let's evaluate the situation separately for the vertical direction and the horizontal direction along the slope.

Considering the direction perpendicular to the slope, two forces are in effect:

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  • The normal force N, directed outward from the slope

Equilibrium exists here, indicating the net force in this direction is zero.

Now let’s examine the parallel direction to the slope. We have two forces present:

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  • The frictional force F_f, acting up the slope

The car moves at a constant speed in this direction, indicating that its acceleration is zero.

a=0

Thus, according to Newton's second law,

F=ma

implying the net force is zero:

F=0

Learn more about slopes and friction:

5 0
21 day ago
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