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Dvinal [7]
3 years ago
8

Data from a survey:

Physics
1 answer:
gayaneshka [121]3 years ago
4 0
C I think the answer is true
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A net force of 50 newtons is applied to a 20 kilogram cart that is already moving at 1 m/s the final speed of the cart was 3 m/s
valina [46]

Answer:

0.8 seconds

Explanation:

F=ma

Let x be the seconds the force is applied.

m = 20kg

F = 50 Newtons (kg*m/sec^2)

acceleration, a, is provided for x seconds to increase the speed from 1 m/s to  3 m/s, an increase of 2m/s

Let's calculate the acceleration of the cart:

F=ma

(50 kg*m/s^2) = (20kg)*a

a = 2.5 m/s^2

---

The acceleration is 2.5 m/s^2.  The cart increases speed by 2.5 m/s every second.  

We want the number of seconds it takes to add 2.0 m/sec to the speed:

(2.5 m/s^2)*x = 2.0 m/s

x = (2.0/2.5) sec

x = 0.8 seconds

7 0
3 years ago
Which steps can be taken to translate the phrase “the height of a tree is increased by seven inches” into an algebraic expressio
ololo11 [35]

Answer:

Explanation:

height of tree = x

seven inches= 7

increaced by=+

3 0
3 years ago
Read 2 more answers
Two light bulbs are wired in series and one bulb burns out (opens.) Technician A says that the other bulb will still work. Techn
Radda [10]

Answer:

Neither technician

Explanation:

Neither technician is correct.

two bulbs are connected in series one bulb burn out  

If one bulb in the series burns out then the circuit will break.

In a series circuit same current passes each resistor.

so, both the technician is incorrect bulb B will not work and current will not increase in the other bulb.

6 0
3 years ago
A proton, which is the nucleus of a hydrogen atom, can be modeled as a sphere with a diameter of 2.4 fm and a mass of 1.67 10-27
Step2247 [10]
For the answer to the question above asking to d<span>etermine the density of the proton. 
</span>Density is mass over volume. 

The volume of a sphere is 4πr³/3. r is half the diameter. 

So the density would be 2.3×10¹⁷ kg/m³. 
I hope my answer helped you. Feel free to ask more questions. Have a nice day!
8 0
4 years ago
Read 2 more answers
A uniformly charged conducting sphere of 1.1 m diameter has a surface charge density of 6.2 µC/m2. (a) Find the net charge on th
ira [324]

Answer:

(a) q = 2.357 x 10⁻⁵ C

(b) Φ = 2.66 x 10⁶ N.m²/C

Explanation:

Given;

diameter of the sphere, d = 1.1 m

radius of the sphere, r = 1.1 / 2 = 0.55 m

surface charge density, σ = 6.2 µC/m²

(a)  Net charge on the sphere

q = 4πr²σ

where;

4πr² is surface area of the sphere

q is the net charge on the sphere

σ is the surface charge density

q = 4π(0.55)²(6.2 x 10⁻⁶)

q = 2.357 x 10⁻⁵ C

(b) the total electric flux leaving the surface of the sphere

Φ = q / ε

where;

Φ is the total electric flux leaving the surface of the sphere

ε is the permittivity of free space

Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)

Φ = 2.66 x 10⁶ N.m²/C

8 0
3 years ago
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