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Barnard’s Star is a red dwarf. It is located 5.9 light years from Earth. (One light year is the same as 9.46 trillion kilometers

mote1985 [20]

A star is located 5.9 light years from Earth.
We know that : 1 light year = 9.46 trillion kilometers.
We will calculate the distance in trillion kilometers multiplying the number of light years by 9.46:
5.9 * 9.46 = 55.814
Answer: The distance is 55.814 trillion km.

5 0

3 years ago

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What is the density (in kg/m3) of a woman who floats in freshwater with 4.92% of her volume above the surface

kipiarov [429]

Answer:

The density of the woman is 950.8 kg/m³

Explanation:

Given;

fraction of the woman's volume above the surface = 4.92%

then, fraction of the woman's volume below the surface = 100 - 4.92% = 95.08%

the specific gravity of the woman $= \frac{95.08}{100 } = 0.9508$

The density of the woman is calculate as;

$Specific \ gravity \ of \ the \ woman = \frac{Density \ of \ the \ woman }{Density \ of \ fresh \ water }\\\\ Density \ of \ the \ woman = Specific \ gravity \ of \ the \ woman \ \times \ Density \ of \ fresh \ water$

Density of fresh water = 1000 kg/m³

Density of the woman = 0.9508 x 1000 kg/m³

Density of the woman = 950.8 kg/m³

Therefore, the density of the woman is 950.8 kg/m³

4 0

2 years ago

A basketball leaves a player's hands at a height of 2.00 m above the floor. The basket is 3.05 m above the floor. The player lik

shutvik [7]

Answer:(10.69, 11.436)

Explanation:

Given

initial height of ball is 2 m

height of basket is 3.05 m

Launching angle $=40^{\circ}$

$x =12\pm 0.27$

y=1.05

equation of trajectory of ball is given by

$y=xtan\theta -\frac{gx^2}{2u^2cos^2\theta }$

for x=12.27

$1.05=12.27\times tan40-\frac{g12.27^2}{2u^2cos^{2}40 }$

u=10.69

for x=11.73

$1.05=11.73\times tan40-\frac{g11.73^2}{2u^2cos^{2}40 }$

u=11.436 m/s

Thus range of speed is (10.69, 11.436)

3 0

2 years ago

When Woods hits a 0.04593 kg golf ball, the golf ball is usually traveling around 281 kilometers per hour. What average force do

Semenov [28]

Answer:

128.9 N

Explanation:

The force exerted on the golf ball is equal to the rate of change of momentum of the ball, so we can write:

$F=\frac{\Delta p}{\Delta t}$

where

F is the force

$\Delta p$ is the change in momentum

$\Delta t=0.030 s$ is the time interval

The change in momentum can be written as

$\Delta p = m(v-u)$

where

m = 0.04593 kg is the mass of the ball

u = 0 is the initial velocity of the ball

$v=281 km/h =78.1 m/s$ is the final velocity of the ball

Substituting into the original equation, we find the force exerted on the golf ball:

$F=\frac{m(v-u)}{\Delta t}=\frac{(0.04593)(78.1-0)}{0.030}=128.9 N$

7 0

3 years ago

A motorboat accelerates uniformly from a velocity of 6.5m/s to the west to a velocity of 1.5m/s to the west. if its accelerate w

grigory [225]

A motorboat accelerates uniformly from a velocity of 6.5m/s to the west to a velocity of 1.5m/s to the west. if its accelerate was 2.7m/s2 to the east , how far did it travel during the accelration? Give your answer in units of kilometers per hour/sec. To find the acceleration of the car we have to 
1. First determine the suitable formula for this word problem.
Which is a. A=vf-vi/t which will be
Given are: Vi= 6.5 m/s Vf= 1.5 m/s a= 2.7 m/sec2 t=1.85s
Solution: 
x = v0t + ½at2
x = 16.645375 m

7 0

3 years ago

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