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snow_lady [41]
3 years ago
10

Let's examine over the next few questions an NFL kick as described in the 3rd Law video. The announcer claimed that the ball is

struck with "almost a ton of force." First of all, phrases like "almost a ton" drive physics people crazy. Give me a number! So let's guess-timate 1900 lbs of force. Convert that to Newtons. (you can always just Google this one)
Physics
1 answer:
hichkok12 [17]3 years ago
4 0

Answer:

8451.62109367671 Newtons

Explanation:

1\ kg=1\times 9.8066500286389=9.8066500286389\ N

1 \lb=2.2046226218488\ kg

It is known that

1\ lbs=\dfrac{9.8066500286389}{2.2046226218488}=4.4482216282508\ N

So,

1900\ lbs=1900\times 4.4482216282508\\\Rightarrow 1900\ lbs=8451.6210936765\ N

The force in Newtons is 8451.6210936765 Newtons

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If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h
raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory (u_x) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

u^2=u_x^2+u_y^2

Where, u_x\ and\ u_y are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1

Now, we know that, for a projectile motion, the maximum height is given as:

H=\frac{u_y^2}{2g}

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m

Therefore, the maximum height reached is 4.63 m.

3 0
3 years ago
Me amor
Oxana [17]

Answer:

jajajaja

Explanation:

7 0
2 years ago
A 5.0-kg box has an acceleration of 2.0 m/s2 when it is pulled by a horizontal force across a surface with μK = 0.50. Find the w
Maslowich

Answer:a) 34.5 N; b) 24.5 N; c) 10 N; d) 1J

Explanation: In order to solve this problem we have to used the second Newton law given by:

∑F= m*a

F-f=m*a where f is the friction force (uk*Normal), from this we have

F= m*a+f=5 Kg*2 m/s^2+0.5*5Kg*9.8 m/s^2= 34.5 N

then f=uk*N=0.5*5Kg*9.8 m/s^2= 24.5N

the net Force = (34.5-24.5)N= 10 N

Finally the work done by the net force is equal to kinetic energy change so

W=∫Force net*dr= 10 N* 0.1 m= 1J

7 0
3 years ago
Read 2 more answers
A container with rigid walls is filled with 4.0 mol of air with Cv=2.5R Then the temperature is increased from 17 degrees C to 3
galina1969 [7]

Explanation:

Internal energy = heat + work

U = Q + W

Since there's no change in volume (rigid walls), W = 0.

U = Q

U = n Cᵥ ΔT

U = (4.0 mol) (2.5 × 8.314 J/mol/K) (354 C − 17 C)

U = 28,000 J

3 0
3 years ago
A ball is dropped from a building of height h. Assume the ball starts from rest and that air friction can be ignored. Derive an
agasfer [191]

Answer:

t=\sqrt{h/g}

Explanation:

We use the kinematics equation to solve this question:

y(t)=y_{o}+v_{o}t+1/2*a*t^{2}

v_{o}=0    because the ball is dropped

a=-g         the acceleration is the gravity, negative because it points downwards

y_{o}=h     initial height

y(t)=h/2     final height

So:

h/2=h-1/2*g*t^{2}

t=\sqrt{h/g}

8 0
3 years ago
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