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Semenov [28]
3 years ago
7

A football quarterback runs 15.0 m straight down the playing field in 2.50 s. he is then hit and pushed 3.00 m straight backward

in 1.75 s. he breaks the tackle and runs straight forward another 21.0 m in 5.20 s. calculate his average velocity (a) for each of the three intervals and (b) for the entire motion.

Physics
1 answer:
My name is Ann [436]3 years ago
3 0
Define
t = time (s),
dt = change in time (s)
 d =distance (m).

The scenario is as follows:
t = 0 s: Begin. Quarterback runs forward.
d = 0

t = 2.5 s: Quarterback is tackled and pushed backward.
d = 15 m

t = 2.5+1.75 = 4.25 s: Quarterback resumes the run forward.
d = 15 - 3 = 12 m

t = 4.25+5.2 = 9.45 s: Quarerback runs another 21 m forward.
d = 12+21 = 33 m

The graph showing the scenario is shown below.

Calculations:
Interval 1:
Average velocty = 15/2.5 = 6.0 m/s

Interval 2:
Average velocity = (12 - 15)/(1.75) = - 1.7 m/s

Interval 3:
Average velocity = (33 - 12)/5.2 = 4.0 m/s

Entire motion:
Average velocity = (33 - 0)/9.45 = 3.5 m/s

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A cannon elevated at 40 degrees is fired at a wall 300m away on level ground. The initial speed of the cannonball is 89m/s How l
lianna [129]
For the answer to the question above, we'll have to use these formulas.

A) to find time to travel the 300m,
just find horizontal component of the velocity and divide.
ie x=89 x t x cos 40, t=x/89 x cos 40 

<span>B) y=vtsin 40 - gt^2/2, just sub in
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I believe you can do the rest.

I hope I helped you with my answers.
3 0
2 years ago
A 180 lb crate is on the ground, and a strong rope is attached. You need to move it across the basement floor, which has a coeff
jekas [21]

Answer:

F = 505.13 N

Yes it is better to pull the rope rather than push it

Explanation:

Let the force is applied at an angle of 60 degree

so we will have net vertical force on the crate is given as

F_n + Fsin60 = mg

here we know

m = 180 lb

m = 81.65 kg

F_n = 81.65(9.81) - Fsin60

F_n = 801 - 0.866 F

now friction force on the crate is given as

F_x = \mu F_n

Fcos60 = 0.7(801 - 0.866 F)

0.5F + 0.61F = 560.7

F = 505.13 N

Yes it is better to pull the rope rather than push it

6 0
3 years ago
What is the moment of inertia of a 2.0 kg, 20-cm-diameter disk for rotation about an axis (a) through the center, and (b) throug
FinnZ [79.3K]

Answer:

(a) I=0.01 kg.m²

(b) I=0.03 kg.m²

Explanation:

Given data

Mass of disk M=2.0 kg

Diameter of disk d=20 cm=0.20 m

To Find

(a) Moment of inertia through the center of disk

(b) Moment of inertia through the edge of disk

Solution

For (a) Moment of inertia through the center of disk

Using the equation of moment  of Inertia

I=\frac{1}{2}MR^{2}\\  I=\frac{1}{2}(2.0kg)(0.20m/2)^{2}\\  I=0.01 kg m^{2}

For (b) Moment of inertia through the edge of disk

We can apply parallel axis theorem for calculating moment of inertia

I=(1/2)MR^{2}+MD\\ Here\\D=R\\I=(1/2)(2.0kg)(0.20m/2)^{2}+(2.0kg)(0.20m/2)^{2}\\  I=0.03kgm^{2}

8 0
3 years ago
Calculate the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m
otez555 [7]

Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

Explanation:

Given: Mass = 5 kg

Spring constant = 6 N/m

Formula used to calculate period is as follows.

T = 2 \pi \sqrt\frac{m}{k}

where,

T = period

m = mass

k = spring constant

Substitute the values into above formula as follows.

T = 2 \pi \sqrt\frac{m}{k}\\= 2 \times 3.14 \times \sqrt\frac{5}{6}\\= 5.73 s

Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

5 0
3 years ago
Why do we learn useless nonsense school, why don't we learn what we will actually use in life
laila [671]
I totally agree but, in my opinion its because of the government and what the state has control over. Teacher have little control over it.
4 0
3 years ago
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