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zaharov [31]
3 years ago
5

Physics question......

Physics
1 answer:
Stella [2.4K]3 years ago
8 0

Answer:

I think it is A or B, but more B as for John used 20 more net force. I recommend doing some research, sorry.

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A car is traveling along a straight road at a velocity of +30.0 m/s when its engine cuts out. For the next 1.79 seconds, the car
Tanzania [10]

Answer:

first value+2nd +3rd

Explanation:

thug life and there

8 0
3 years ago
A pencil has a density of 0.875g/ml. It has a mass of 3.50 g. What is the volume<br><br>​
erica [24]

Answer:

V = 4 cm^3

Explanation:

Divide the mass by the density.

3.50g/0.875g/mL = 4cm^3

7 0
3 years ago
Q11) If you were standing at the top of a building and you dropped a rock.
Dafna1 [17]

Answer:

Part A

The distance travel by the rock is approximately 132.496 m

Part B

The speed when the rock hits the ground is approximately 50.96 m/s

Explanation:

Part A

The question is focused on the kinetics equation of a free falling object

The given parameter is the time it takes the rock to hit the ground, t = 5.2 s

For an object in free fall, we have;

h = 1/2·g·t²

Where;

h = The height from which the object is dropped

g = The acceleration due to gravity ≈ 9.8 m/s²

t = The time taken to travel the distance, h = 5.2 s

∴ h = 1/2 × 9.8 m/s² × (5.2 s)² ≈ 132.496 m

The distance travel by the rock, h ≈ 132.496 m

Part B

The speed, 'v', when the rock hits the ground, is given by the following kinematic equation,

v = g·t

∴ v = 9.8 m/s² × 5.2 s = 50.96 m/s

The speed when the rock hits the ground, v ≈ 50.96 m/s.

8 0
3 years ago
The archerfish uses a remarkable method for catching insects sitting on branches or leaves above the waterline. The fish rises t
Degger [83]

Answer:

The height above the waterline that the stream reaches at the horizontal position of the insect is 15 cm.

Explanation:

Please, see the attached figure for a description of the problem.

The motion is parabolic and this is the equation that describes the position of an object in such a motion:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α  + 1/2 · g · t²)

Where:

r = position vector

x0 = initial horizontal position

v0 = initial velocity

t = time

α = angle of the stream with the waterline

y0 = initial vertical position

g = acceleration due to gravity

First, let´s calculate how much time it takes the stream to reach the horizontal position of 0.27 m. For this, we will use the equation of the x-component of the vector position:

x = x0 + v0 · t · cos α

Since the origin of the reference system is located at the mouth of the fish, x0 = 0. Then:

0.27 m = 3.7 m/s · t · cos 35º

t = 0.27 m /(3.7 m/s · cos 35º)

t = 0.089 s

Now, with this time, we can calulate the vertical position (height) of the stream using the equation for the y-component of the vector "r":

y = y0 + v0 · t · sin α  + 1/2 · g · t²

y = 0 m + 3.7 m/s · 0.089 s · sin 35º + 1/2 · (-9.8 m/s²) · (0.089s)²

y = 0.15 m

when the stream reaches 27 cm horizontally, it will reach 15 cm vertically and hit the insect!

3 0
4 years ago
T=2pi square root 1/g solve for g.<br> Explanation would be really helpful.
Natalija [7]

I added individual steps for clarity. Note that g must be positive if the solution is to be real.

T=2\pi \sqrt{\frac{1}{g}}=2\pi g^{-\frac{1}{2}}\\g^{-\frac{1}{2}} = \frac{T}{2\pi}\\(g^{-\frac{1}{2}})^{-2} = (\frac{T}{2\pi})^{-2}\\g = \frac{4\pi^2}{T^2}\,\,\,, g>0}

Let me know if you have any questions.

7 0
3 years ago
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