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nexus9112 [7]
3 years ago
12

Ryan wants to use the fermi process to estimate the number of basketballs that would fill a gymnasium. as part of the process, h

e needs to estimate the volume of a basketball. the diameter of a basketball is about 10 in. which estimate best approximates the volume of a basketball? 5×101 in³ 5×102 in³ 5×103 in³ 5×104 in³ brainly
Physics
2 answers:
vredina [299]3 years ago
8 0

Answer:

its 5×102 in³

Explanation:

i just took the test for it

GREYUIT [131]3 years ago
7 0
The answer is 5×102 in³

You might be interested in
light of a certain frequency has a wavelength of 438 nm in water.What is the wavelength of this light in benzene​
sveticcg [70]

Answer:

388.97 nm

Explanation:

The computation of the wavelength of this light in benzene is shown below:

As we know that

n (water) = 1.333

n (benzene) = 1.501

\lambda (water) \times n(water) = \lambda (benzene) \times n(benzene)

And, the wavelength of water is 438 nm

\lambda (benzene) = \lambda (water) [\frac{n(water)}{n(benzene}]

Now placing these values to the above formula

So,

= 438 \times \frac{1.333}{1.501}

= 388.97 nm

We simply applied the above formula so that we can easily determine the wavelength of this light in benzene​ could come

5 0
3 years ago
If the potential difference across a 12-ohm resistor is 6 volts, the current through the resistor is?
larisa86 [58]
Current is inversely proportional to the resistance of the resistor and directly to the potential difference across it.

I = V/R = 6/12 = 0.5 A
4 0
3 years ago
The question is in the picture
Sedbober [7]

Answer:

e) 120m/s

Explanation:

When the ball reaches its highest point, its velocity becomes zero, meaning

v_0-gt = 0.

where v_0 is the initial velocity.

Solving for t we get

t = \dfrac{v_0}{g}

which is the time it takes the ball to reach the highest point.

Now, after the ball has reached its highest point, it turns around and falls downwards. After time t_0 since it had reached the highest point, the ball has traveled downwards and the velocity v_f it has gained is

v_f = gt_0,

and we are told that this is twice the initial velocity v_0; therefore,

v_f = 2v_0  = gt_0

which gives

t_0 = \dfrac{2v_0}{g}.

Thus, the total time taken to reach velocity 2v_0 is

t_{tot} = t+t_0 = \dfrac{v_0}{g}+\dfrac{2v_0}{g}

t_{tot} = \dfrac{3v_0}{g}.

This t_{tot}, we are told, is 36 seconds; therefore,

36= \dfrac{3v_0}{g},

and solving for v_0 we get:

v_0 = \dfrac{36g}{3}

v_0 = \dfrac{36s(10m/s^2)}{3}

\boxed{v_0 = 120m/s}

which from the options given is choice e.

7 0
3 years ago
While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)
ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration=2.40m/s^{2}, c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = 2.40m/s^{2}

time = 12.0 s

initial position = 0

c)

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}

after cancelling the pertinent units, I get that  my answer will be given in meters. So I get:

x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}

which solves to:

x=172.8m

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

a=\frac{V_{f}-V_{0} }{t}

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

at=V_{f}-V_{0}

and finally I can add V_{0} to both sides so I get:

V_{f}=at+V_{0}

and now I can proceed and substitute the known values:

V_{f}=at+V_{0}

V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0

which solves to:

V_{f}=28.8m/s

8 0
3 years ago
Read 2 more answers
A traves de una manguera de 1 in de diámetro fluye gasolina con una velocidad media de 5ft/s ¿cuál es el gasto?
jonny [76]

Answer:

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

Explanation:

El gasto es el flujo volumétrico de gasolina (Q), medido en pies cúbicos por segundo, que sale de la manguera. Asumiendo que la velocidad de salida es constante, tenemos que el gasto a través de la manguera es:

Q = \frac{\pi}{4}\cdot D^{2}\cdot v (1)

Donde:

D - Diámetro de la manguera, medido en pies.

v - Velocidad medida de salida, medida en pies por segundo.

Si sabemos que D = \frac{1}{12}\,ft y v = 5\,\frac{ft}{s }, entonces el gasto de gasolina es:

Q = \frac{\pi}{4}\cdot \left(\frac{1}{12}\,ft \right)^{2} \cdot \left(5\,\frac{ft}{s} \right)

Q \approx 0.0273\,\frac{ft^{3}}{s}

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

6 0
3 years ago
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