If bert the bat traveles eastward at 40 mph with a tail wind of 6 mph,what is his actual speed?

Which of the following is a strength training option?

Vilka [71]

Answer:

D. All of the above

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5 0

2 years ago

Read 2 more answers

B) Calculate the net work required to bring a 1300 kg car moving at 30 m/s to rest.

prohojiy [21]

I Lolo my name is keshav and

5 0

2 years ago

The intensity of light from a star (its brightness) is the power it outputs divided by the surface area over which it’s spread:

kow [346]

Answer:

$\frac{d_{1}}{d_{2}}=0.36$

Explanation:

1. We can find the temperature of each star using the Wien's Law. This law is given by:

$\lambda_{max}=\frac{b}{T}=\frac{2.9x10^{-3}[mK]}{T[K]}$ (1)

So, the temperature of the first and the second star will be:

$T_{1}=3866.7 K$

$T_{2}=6444.4 K$

Now the relation between the absolute luminosity and apparent brightness is given:

$L=l\cdot 4\pi r^{2}$ (2)

Where:

L is the absolute luminosity
l is the apparent brightness
r is the distance from us in light years

Now, we know that two stars have the same apparent brightness, in other words l₁ = l₂

If we use the equation (2) we have:

$\frac{L_{1}}{4\pi r_{1}^2}=\frac{L_{2}}{4\pi r_{2}^2}$

So the relative distance between both stars will be:

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{L_{1}}{L_{2}}$ (3)

The Boltzmann Law says, $L=A\sigma T^{4}$ (4)

σ is the Boltzmann constant
A is the area
T is the temperature
L is the absolute luminosity

Let's put (4) in (3) for each star.

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{A_{1}\sigma T_{1}^{4}}{A_{2}\sigma T_{2}^{4}}$

As we know both stars have the same size we can canceled out the areas.

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{T_{1}^{4}}{T_{2}^{4}}$

$\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}$

$\frac{d_{1}}{d_{2}}=0.36$

I hope it helps!

5 0

3 years ago

A student throws a ball upward with a velocity of 35 m/s. What is the acceleration of the ball as it rises to the top of its arc

lora16 [44]

Answer:

The acceleration of the ball as it rises to the top of its arc equals 9.807 meters per square second.

Explanation:

Let suppose that maximum height of the arc is so small in comparison with the radius of the Earth.

Since the ball is launched upwards, then the ball experiments a free-fall motion, that is, an uniform accelerated motion in which the element is accelerated by gravity. Then, the acceleration experimented by the motion remains constant at every instant and position.

Besides, the gravitational acceleration in the Earth and, in consequence, the acceleration of the ball as it rises to the top of its arc equals 9.807 meters per square second.

7 0

2 years ago

What is the minimum speed needed to fire a champagne cork a distance of 11m?

Crank

To start with solving this problem, let us assume a launch angle of 45 degrees since that gives out the maximum range for given initial speed. Also assuming that it was launched at ground level since no initial height was given. Using g = 9.8 m/s^2, the initial velocity is calculated using the formula:

(v sinθ)^2 = (v0 sinθ)^2 – 2 g d

where v is final velocity = 0 at the peak, v0 is the initial velocity, d is distance = 11 m

Rearranging to find for v0: <span>
v0 = sqrt (d * g/ sin(2 θ)) </span>

8 0

3 years ago