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lisabon 2012 [21]

2 years ago

4

A student eats a dinner rated at 2000 (food) Calories. He wishes to do an equivalent amount of work in the gymnasium by lifting

a 50 kg mass. How many times must he raise the mass to expend this amount of energy? Assume that he raises it a distance of 2 m each time and that he regains no energy when it is dropped to the floor. The acceleration due to gravity is 9.8 m/s2 and 1 food Calorie is 103 calories.

1 answer:

Morgarella [4.7K]2 years ago

6 0

Answer:

879.49

Approximately 880 times

Explanation:

2000 (food) Calories = 2000 * 103 calories

2000 (food) Calories = 206000 cal.

Work done by the boy is given by Mgh , potential work

Work= 50*2*9.8

Work = 980 joules.

1 kilocalorie =

4184 joules

So 1 cal = 4.185 joules

206000 calories = 206000*4.184

206000 cal = 861904 joules.

To extend 206000 cal,

He will lift the 50 kg mass 861904/980 times

= 879.49 times.

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Aleonysh [2.5K]

Answer: $53.31\°$ East of North

Explanation:

We have the following data:

Speed of the wind from East to West: $6.68 m/s$

Speed of the bee relative to the air: $8.33 m/s$

If we graph these speeds (which in fact are velocities because are vectors) in a vector diagram, we will have a right triangle in which the airspeed of the bee (its speed relative to te air) is the hypotense and the two sides of the triangle will be the <u>Speed of the wind from East to West</u> (in the horintal part) and the <u>speed due North relative to the ground</u> (in the vertical part).

Now, we need to find the direction the bee should fly directly to the flower (due North):

$sin \theta=\frac{Windspeed-from-East-to-West}{Speed-bee-relative-to-air}$

$sin \theta=\frac{6.68 m/s}{8.33 m/s}$

Clearing $\theta$ :

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6 0

2 years ago

To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th

Kryger [21]

Answer:

$H = 10.05 m$

Explanation:

If the stone will reach the top position of flag pole at t = 0.5 s and t = 4.1 s

so here the total time of the motion above the top point of pole is given as

$\Delta t = 4.1 - 0.5 = 3.6 s$

now we have

$\Delta t = \frac{2v}{g}$

$3.6 = \frac{2v}{9.8}$

$v = 17.64 m/s$

so this is the speed at the top of flag pole

now we have

$v_f - v_i = at$

$17.64 - v_i = (-9.8)(0.5)$

$v_i = 22.5 m/s$

now the height of flag pole is given as

$H = \frac{v_f + v_i}{2}t$

$H = \frac{22.5 + 17.64}{2} (0.5)$

$H = 10.05 m$

5 0

2 years ago

What is the speed of a beam of electrons when the simultaneous influence of an electric field of 1.56×104v/m and a magnetic fiel

sashaice [31]

1) $3.38\cdot 10^6 m/s$

When both the electric field and the magnetic field are acting on the electron normal to the beam and normal to each other, the electric force and the magnetic force on the electron have opposite directions: in order to produce no deflection on the electron beam, the two forces must be equal in magnitude

$F_E = F_B\\qE = qvB$

where

q is the electron charge

E is the magnitude of the electric field

v is the electron speed

B is the magnitude of the magnetic field

Solving the formula for v, we find

$v=\frac{E}{B}=\frac{1.56\cdot 10^4 V/m}{4.62\cdot 10^{-3} T}=3.38\cdot 10^6 m/s$

2) 4.1 mm

When the electric field is removed, only the magnetic force acts on the electron, providing the centripetal force that keeps the electron in a circular path:

$qvB=m\frac{v^2}{r}$

where m is the mass of the electron and r is the radius of the trajectory. Solving the formula for r, we find

$r=\frac{mv}{qB}=\frac{(9.1 \cdot 10^{-31} kg)(3.38\cdot 10^6 m/s)}{(1.6\cdot 10^{-19} C)(4.62\cdot 10^{-3}T)}=4.2\cdot 10^{-3} m=4.1 mm$

3) $7.6\cdot 10^{-9}s$

The speed of the electron in the circular trajectory is equal to the ratio between the circumference of the orbit, $2 \pi r$ , and the period, T:

$v=\frac{2\pi r}{T}$

Solving the equation for T and using the results found in 1) and 2), we find the period of the orbit:

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2 years ago

A roller coaster, traveling with an initial speed of 15 meters per second, decelerates uniformly at â7.0 meters per second2 to a

Harrizon [31]

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2 years ago

A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second

Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

$v_x = 15 m/s$

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

$d= v_x t =(15 m/s)(5 s)=75 m$

5 0

2 years ago