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Talja [164]
2 years ago
14

A cheerleader lifts his 79.4 kg partner straight up off the ground a distance of 0.945 m before releasing her. the acceleration

of gravity is 9.8 m/s 2 . if he does this 33 times, how much work has he done? answer in units of j
Physics
1 answer:
Oksi-84 [34.3K]2 years ago
5 0
To find out how much work he has done, we must first calculate force using the force formula (F= Mass*Acceleration). In this case, mass is 79.4 and acceleration is the gravitational constant of 9.8m/s, plugging this into the formula we find that force is 778.12Newtons. Next, we need to multiply force by the distance to get the amount of energy used to lift his partner once. Which is 778.12 * .945 = 735.32. Finally, we need to multiply 735.32 by the number of times he lifts his partner, 33, to get 735.32 * 33 to find that the energy he has expended 24,265.56 Joules of energy.
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3. If a car is moving at 90km/hr and it rounds a corner, also at 90km/hr. Does it maintain
dolphi86 [110]

Answer:

Constant speed: yes

Constant velocity: no

Explanation:

Let's remind the definition of speed and velocity:

- Speed is a scalar quantity, which is equal to the ratio between the distance covered (regardless of the direction) and the time taken:

s=\frac{d}{t}

- Velocity is a vector quantity, so it has both a magnitude and a direction. The magnitude is equal to the rate between the displacement of the object and the time taken, while the direction is the same as the displacement.

In this problem, we notice that:

- The speed of the car remains constant, as it is 90 km/h

- However, its direction of motion changes while the car travels round the corner: this means that the direction of the velocity is also changing, therefore velocity is not constant.

8 0
3 years ago
Give reasons.<br>A stone thrown upward returns back to the earth surface.​
Wittaler [7]

Answer:

Gravity

Explanation:

Due to earths gravity, anything that goes into the air returns back to the surface unless it is given the ability to fly.

Hope this helps!

7 0
3 years ago
An astronaut is in space with a baseball and a bowling ball. The astronaut pushes both objects in the same direction. If both ba
Crazy boy [7]

Answer:

As a mass greater than that of baseball, at the moment of the bowling wave the moment of the baseball ball is also greater

Explanation:

This problem is an application of momentum and momentum. When the astronaut pushed balls, he needed more force to move the ball of greater mass (bowling). The expression for soul is

      p = m v

Besibol Blade

      p1 = m1 v

Bowling ball

      p2 = m2 v

As a mass greater than that of baseball, at the moment of the bowling wave the moment of the baseball ball is also greater

      p2 >> p1

3 0
3 years ago
A discharge lamp rated at 25 W (1 W = 1 J/s) emits yellow light of wavelength 580 nm. How many photons of yellow light does the
Scrat [10]

Answer:

the number of photons of yellow light does the lamp generate in 1.0 s is 7 x 10^{19}

Explanation:

given information:

power, P = 25 W

wavelength. λ - 580 nm = 5.80 x 10^{-7} m

time, t = 1 s

to calculate the number of photon(N), we use the following equation

N = λPt/hc

where

λ = wavelength (m)

P = power (W)

t = time interval (s)

h = Planck's constant (6.23 x 10^{-34} Js)

c = light's velocity (3 x 10^{8} m/s^{2})

So,

N = λPt/hc

   = (5.80 x 10^{-7})(25)(1)/(6.23 x 10^{-34})(3 x 10^{8} m/s^{2})

    = 7 x 10^{19}

3 0
2 years ago
A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp
NISA [10]

A) 750 m

First of all, let's find the wavelength of the microwave. We have

f=12GHz=12\cdot 10^9 Hz is the frequency

c=3.0\cdot 10^8 m/s is the speed of light

So the wavelength of the beam is

\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

y=\frac{m\lambda D}{a}

where

m = 1 since we are interested only in the central fringe

D = 30 km = 30,000 m

a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)

Substituting, we find

y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m

and so, the diameter is

d=2y = 750 m

B) 0.23 W/m^2

First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

r=\frac{750 m}{2}=375 m

So the area is

A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2

And since the power is

P=100 kW = 1\cdot 10^5 W

The average intensity is

I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2

4 0
3 years ago
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