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3 years ago

Just need to know net force, unbalanced or balanced and accelerated or not accelerated :)

Physics

1 answer:

Annette [7]3 years ago

5 0

Answer:

9. forces: BALANCED, object will: NOT ACCELERATE

10. Forces: UNBALANCED, object will: ACCELERATE

Explanation: sorry I don't no the net force but hope this helps for know

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help me pass the test please there is a couple of questions in my profile to answer i am leaving this one for you to help

kompoz [17]

Answer:

0.5 m/s².

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 10 m/s

Time (t) = 20 s

Acceleration (a) =?

Acceleration can simply be defined as the rate of change of velocity with time. Mathematically, it is expressed as:

a = (v – u) / t

Where:

a is the acceleration.

v is the final velocity.

u is the initial velocity.

t is the time.

With the above formula, we can obtain the acceleration of the car as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 10 m/s

Time (t) = 20 s

Acceleration (a) =?

a = (v – u) / t

a = (10 – 0) / 20

a = 10/20

a = 0.5 m/s²

Therefore, the acceleration of the car is 0.5 m/s².

6 0

3 years ago

The sound level measured in a room by a person watching a movie on a home theater system varies from 40 dB during a quiet part t

UNO [17]

Answer:

$\alpha=-3.01dB$

Explanation:

From the question we are told that:

Sound level intensity

$\triangle I=40dB-80dB$

Generally the equation for intensity level is mathematically given by

$\alpha=10log_{10}(I/I_x)dB$

Where

I= Intensity measured

$I_x=Threshold\ of\ audibility$

$I_x= 10-12 W / m2$

$\alpha= 10 log10 \frac{I_1}{I_x} - 10 log10 \frac{}I_2{I_x}$

$\alpha= 10 log10 \frac{I_1}{I_2}$

$\alpha=10 log10\frac{40}{80}$

$\alpha=-3.01dB$

7 0

3 years ago

Assuming there are no accidents or delays, the distance that a car travels down the interstate

Ganezh [65]

Explanation:

The distance that a car travels down the interstate can be calculated with the following formula:

Distance = Speed x Time

(A) Speed of the car, v = 70 miles per hour = 31.29 m/s

Time, d = 6 hours = 21600 s

Distance = Speed x Time

D = 31.29 m/s × 21600 s

D = 675864 meters

$D=6.75\times 10^5\ m$

(b) Time, d = 10 hours = 36000 s

Distance = Speed x Time

D = 31.29 m/s × 36000 s

D = 1126440 meters

$D=1.12\times 10^6\ m$

(c) Time, d = 15 hours = 54000 s

Distance = Speed x Time

D = 31.29 m/s × 54000 s

D = 1689660 meters

$D=1.68\times 10^6\ m$

Hence, this is the required solution.

7 0

3 years ago

The cold virus causes disease when it enters the _____ of the human body.

belka [17]

Cells of the human body

7 0

4 years ago

Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi

nadya68 [22]

Answer:

$E = 1.85*10^{12}\frac{N}{C}$

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction, and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

$E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}$

Then the electric field at the point of interest is estimated as:

$E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}$

6 0

4 years ago