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Tatiana [17]
3 years ago
7

a defensive tackle picks up the 0.5kg football to a height of 0.8m in 0.25s .... 1.calculate the work done 2.calculate the power

generated
Physics
1 answer:
IrinaVladis [17]3 years ago
7 0
1).  The work done is the potential energy the football has after it is raised.

Weight of the football = (Mass) x (gravity) = (0.5) x (9.8) = 4.9 newtons

Work= (force) x (distance) = (4.9 newtons) x (0.8 meter) = <em>3.92 joules</em>

2). Power = (work) / (time)  (You know this from the car problem.)

Power = (3.92 joules) / 0.25 s) = 15.68 joules per second = <em>15.68 watts</em>.
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You're driving down the highway late one night at 20 m/s when a deer steps onto the road 49 m in front of you. You reaction time
Leno4ka [110]

Answer:

v = 26.7 m/s

Explanation:

Given,

speed of the car = 20 m/s

distance between the car and the deer = 49 m

time taken to press the brake = 0.50 s

maximum deceleration of the car = 10 m/s²

Now,

distance travel by the car in 0.5 s = u x t = 20 x 0.5 = 10 m

distance travel by the car after the break is pressed

Using equation of motion

v² = u² + 2 a s

0² = 20² - 2 x 10 x s

s = 20 m

Total distance travel by the car = 20 + 10 = 30 m

Distance between deer and car = 49-30 = 19 m.

b. Maximum speed a car could have

Distance travel by the car in reaction time = v' x 0.5

v' is the maximum speed of the car.

maximum distance car can cover = 49 - 0.5 v'

Now, Using equation of motion

v² = u² + 2 a s

0² =v'² - 2 x 10 x (49- 0.5 x v')

v'² +10 v' -980 = 0

By solving

v = 26.7 m/s

Hence, maximum speed of the car can be 26.7 m/s

 

4 0
3 years ago
A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm
Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

a)

  • In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

       F = - k * \Delta x (1)

  • where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
  • Solving for k:

       k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)

b)

  • Assuming no friction present, total mechanical energy mus keep constant.
  • When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

        U = \frac{1}{2}* k* (\Delta x)^{2} (3)

  • Replacing k and Δx by their values, we get:

       U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)

c)

  • When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
  • We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
  • So, we can write the following expression:

        \frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2}  = \frac{1}{2}*k*\Delta x^{2}   (5)

  • Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

        \frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2}  = 4J   (6)

⇒      \frac{1}{2}* 200N/m* \Delta x_{1} ^{2}   = 4J  - 3.6 J = 0.4 J (7)

⇒     \Delta x_{1}   = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)

6 0
2 years ago
Help please i will mark brainlist!!
Dmitrij [34]
The answer is A.


p=m/v
p= 240/60
p= 4 g/cm^3
7 0
3 years ago
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A small 20-kg canoe is floating downriver at a speed of 2 m/s. What is the canoe’s kinetic energy? A. 40 J B. 80 J C. 18 J
rusak2 [61]

A small 20-kg canoe is floating downriver at a speed of 2 m/s. 40 J is the canoe’s kinetic energy.

Answer: Option A

<u>Explanation:</u>

The given canoe has the mass and is being given to move at a speed. Therefore the kinetic energy of the canoe can be calculated using the following method,

Given that mass of the canoe = 20 kg and its speed =1 m/s

As we know that the Kinetic energy has the formula,

\text {Kinetic energy}=\frac{1}{2} \boldsymbol{m} \boldsymbol{v}^{2}

Therefore, substituting the value into the equation, we get,  

K . E .=\frac{1}{2} \times 20 \times 2^{2} = 40 J

4 0
3 years ago
Read 2 more answers
Calculate the speed of the car at each checkpoint by
lana66690 [7]

Answer: The speed at the first quarter checkpoint is 0.74 m/s. The speed at the second quarter checkpoint is 1.40 m/s. The speed at the third quarter checkpoint is 1.61 m/s. The speed at the finish line is 1.89 m/s.

Explanation: I did the assignment and got it correct :)

3 0
2 years ago
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