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3 years ago

I need help with this one 2 if you guys don’t mind

Physics

1 answer:

zloy xaker [14]3 years ago

6 0

3 is 3.81 meters

4 is 0.47 liters

5 is 4 cm

6 is 23 mm

7 is 53 m

8 is 1800 mg

9 is 31.07 mi

Hope I’m helping ya

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How long does it take light to travel 850 km in a vacuum? Answer in ms.(Express your answer to two significant figures.)

poizon [28]

Answer:

0.002833 sec

Explanation:

Speed of light in vacuum is $3\times 10^{8}m/sec$

Given distance = 850 km = 850×1000=850000 m

We have to calculate the time that light take to travel the distance 850 km

Time $T=\frac{distance }{speed}=\frac{850000}{3\times 10^8}=2.833\times 10^{-3}sec$

So the time taken by light to travel 850 km is 0.002833 sec

5 0

3 years ago

A bullet with an initial kinetic energy of 400 J strikes a wooden block where a 8000 N resistive force stops the bullet. What is

Sindrei [870]

Answer:

d = 0.05 [m] = 50 [mm]

Explanation:

We must remember the principle of conservation of energy which tells us that energy is transformed from one way to another. For this case, the initial kinetic energy is transformed into useful work that is equal to the product of force by distance.

$E_{k}=F*d\\400 = 8000*d\\d = 0.05 [m] = 50 [mm]$

5 0

2 years ago

Suppose that you are headed toward a plateau 55 meters high. If the angle of elevation to the top of the plateau is 40degrees,

Iteru [2.4K]

Answer:

$x=65.55m$

Explanation:

Let x be the distance to the shore

From trigonometry properties:

$tan(40^{o} )=\frac{55m}{x} \\x=\frac{55m}{tan(40^{o} )} \\x=65.55m$

3 0

3 years ago

A diver springs upward from a board that is 2.70 m above the water. At the instant she contacts the water her speed is 10.9 m/s

TiliK225 [7]

Answer:

vo=5.87m/s

Explanation:

Hello! In this problem we have a uniformly varied rectilinear movement.

Taking into account the data:

α =69.2

vf = 10m / s

h=2.7m

g=9.8m/s2

We know we want to know the speed on the y axis.

We calculate vfy

vfy = 10m / s * (sen69.2) = 9.35m / s

We can use the following equation.

$vf^{2} =vo^{2}+2*g*h\\$

We clear the vo (initial speed)

$vo=\sqrt{vf^{2}-2*g*h }$

$v0=\sqrt{(9.35m/s)^{2}-2*9.8m/s^{2} *2.7m}$

vo=5.87m/s

7 0

2 years ago

Objects with masses of 120 kg and a 420 kg are separated by 0.380 m. (a) Find the net gravitational force exerted by these objec

SOVA2 [1]

Answer:

$F_{net} = 6.879\times 10^{- 7}\ N$

Solution:

As per the question:

Mass of first object, m = 120 kg

Mass of second object, m' = 420 kg

Mass of the third object, M = 69.0 kg

Distance between the m and m', d = 0.380 m

Now,

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m:

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F = \frac{6.67\times 10^{-11}\times 120\times 0.69}{\frac({0.380}{2}^{2})} = 1.529\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m':

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F' = \frac{6.67\times 10^{-11}\times 420\times 0.69}{\frac({0.380}{2}^{2})} = 5.35\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m and m':

$F_{net} = F + F'$

$F_{net} = 1.529\times 10^{- 7} + 5.35\times 10^{- 7} = 6.879\times 10^{- 7}\ N$

6 0

3 years ago