Blank objects is a major cause of back injuries in the workplace

After hitting a long fly ball that goes over the right fielder's head and lands in the outfield, the batter decides to keep goin

jarptica [38.1K]

Answer:

a) 633.39 J

b) 0.28

Explanation:

a)The kinetic energy of the player = $\frac{1}{2} mv^{2}$

Work done by friction = energy change of the player

= $\frac{1}{2} 67(4.35)^{2}$ = 633.9 J

b) Assuming the frictional force stays constant,

Work done by friction = Frictional force×distance

Frictional force = kinetic friction(μ)×normal reaction

Normal reaction = weight = mass×gravitational acceleration ( g=10m/s2 )

Combining these equations

633.9 = F×3.4 ⇒ F = 186.44 N

F = μmg ⇒ μ = F/mg

= 186.44/670

= 0.28

5 0

3 years ago

A proton is ejected from the sun at a speed of 2 x 10^6 m/s. How long does it take for this proton to reach earth? Answer in hou

wolverine [178]

The distance from the Sun to the Earth is 149,600,000 km.

$d=149 600 000 \ km = 149 600 000 000 \ m = 1496 \cdot 10^8 \ m \\
v=2 \cdot 10^6 \ \frac{m}{s} \\ \\ \\
t=\frac{d}{v} \\ \\
t=\frac{1496 \cdot 10^8 \ m}{2 \cdot 10^6 \ \frac{m}{s}}=748 \cdot 10^2 \ s =\frac{748 \cdot 10^2 }{36 \cdot 10^2} \ h \approx \underline{\underline{20,78 \ h}}$

5 0

3 years ago

How fast would the car need to go to double its kinetic energy?

boyakko [2]

Answer:

$v_{f} = \sqrt{2}\cdot v_{o}$

Explanation:

Let consider a car travelling at a speed $v_{o}$ . The ratio of final kinetic energy to initial kinetic energy:

$\frac{\frac{1}{2}\cdot m \cdot v^{2}_{f} }{\frac{1}{2}\cdot m \cdot v^{2}_{o}} = 2$

$\frac{v_{f}}{v_{o}} = \sqrt{2}$

The final speed is:

$v_{f} = \sqrt{2}\cdot v_{o}$

3 0

3 years ago

Read 2 more answers

When you exert 75 N on a jack to lift a 6000 N car, what is the jack’s actual mechanical advantage? Show your work.

professor190 [17]

Answer:

80

Explanation:

the mechanical advantage is the ratio of the load to the effort so it doesn't have units.to calculate it you use the formula

mechanical advantage=load/effort

in this case the load is 6000N and the effort is 75N

Ma=6000/75

=80

I hope this helps

3 0

3 years ago

Suppose an oven's radiation wavelength is 0.125 m. a container with 350.00 g of water was placed in the oven, and the temperatur

ivann1987 [24]

The heat (energy) needed to raise the temperature of the water is given by
$Q=m C_S (T_f - T_i)=(350.0 g)(4.18 J/gC)(80C-20C)=87780 J$

The wavelength of the radiation of the oven is $\lambda=0.125 m$ , so the energy of a single photon of this radiation is
$E=h \frac{c}{\lambda}=(6.6 \cdot 10^{-34}J) \frac{3\cdot 10^8 m/s}{0.125 m}=1.6 \cdot 10^{-24} J$

So, the number of photons required to heat the water is the total energy absorbed by the water divided by the energy of a single photon:
$N= \frac{Q}{E}= \frac{87780 J}{1.6\cdot 10^{-24}J}= 5.5 \cdot 10^{28}$ photons

8 0

3 years ago