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g100num [7]
3 years ago
10

Which describes acceleration involving only a change in direction?

Physics
2 answers:
Shkiper50 [21]3 years ago
6 0
B, since it includes both an increase in speed and a change in direction.
elena-14-01-66 [18.8K]3 years ago
6 0

Answer is A. moving at constant speed around a circular track

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PLEASE HELP!
lana66690 [7]

Answer:

x = 4.32 [m]

Explanation:

We must divide this problem into three parts, in the first part we must use Newton's second law which tells us that the force is equal to the product of mass by acceleration.

∑F = m*a

where:

F = force = 700 [N]

m = mass = 2030 [kg]

a = acceleration [m/s²]

Now replacing:

F=m*a\\700=2030*a\\a = 0.344[m/s^{2}]

Then we can determine the final speed using the principle of conservation of momentum and amount of movement.

(m_{1}*v_{1})+Imp_{1-2}=(m_{1}*v_{2})

where:

m₁ = mass of the car = 2030 [kg]

v₁ = velocity at the initial moment = 0 (the car starts from rest)

Imp₁₋₂ = The impulse or momentum (force by the time)

v₂ = final velocity after the impulse [m/s]

(2030*0) + (700*5)=(2030*v_{2})\\3500 = 2030*v_{2}\\v_{2}=1.72[m/s]

Now using the following equation of kinematics, we can determine the distance traveled.

v_{2}^{2} =v_{1}^{2}+2*a*x

where:

v₂ = final velocity = 1.72 [m/s]

v₁ = initial velocity = 0

a = acceleration = 0.344 [m/s²]

x = distance [m]

1.72^{2}=0^{2} +(2*0.344*x) \\2.97 = 0.688*x\\x = 4.32 [m]

8 0
3 years ago
The heliocentric theory was first proposed by the astronomer
Setler [38]
Nicholas Copernicus, Greek astronomer, was the one who famously proposed the heliocentric theory -- the theory by which the sun was in the center of the solar system and the planets orbited it circularly.

Hope I helped and good luck!
5 0
4 years ago
Water is flowing through a channel that is 12m wide with a
Alexus [3.1K]

Answer:

Velocity from second channel will be 1.6875 m/sec

Explanation:

We have given width of the channel , that is diameter of the channel 1 d_1 = 12 m

So radius r_1=\frac{d_1}{2}=\frac{12}{2}=6m

Speed through the channel 1 v_1=0.75m/sec

Width , that is diameter of the channel 2 d_2=4m

So r_2=2m

From continuity equation

A_1v_1=A_2v_2

\pi \times 12^2\times 0.75=4\times \pi\times  4^2\times v_2

v_2=1.6875m/sec

So velocity from smaller channel will be 1.6875 m /sec

6 0
3 years ago
What is the electric field strength just outside the flat surface of the conductor?
inna [77]
We can find the answer step-by-step:

1) The electric charges on a conductor must lie entirely on its surface. This is because the charges have same sign, so the force acting between each other is repulsive therefore the charges must be as far apart as possible, i.e. on the surface of the conductor.

2) We consider a cylinder perpendicular to the surface of the conductor, that crosses the surface with its section. We then apply Gauss law, which states that the flux of the electric field through this cylinder is equal to the total charge inside it divided the electrical permittivity:
\Phi =  \frac{Q}{\epsilon_0}

3) The electric field outside the surface is perpendicular to the surface itself (otherwise there would be a component of the electric force parallel to the surface, which would move the charge, violating the condition of equilibrium). The electric field inside the conductor is instead zero, because otherwise charges would move violating again equilibrium condition. Therefore, the only flux is the one crossing the section A of the cylinder outside the surface: 
\Phi = E A

4) The total charge contained in the cylinder is the product between the section, A, and the charge density \sigma on the surface of the conductor:
Q=\sigma A

5) Substituting the flux and the charge density inside Gauss law, we can find the electric field just outside the surface of the conductor:
EA= \frac{\sigma A}{\epsilon_0}
therefore
E= \frac{\sigma}{\epsilon_0}
4 0
4 years ago
How to do number 19?
ivann1987 [24]
The answer is 8/3. You have to simplify each side and isolate the variable. 
5 0
3 years ago
Read 2 more answers
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