All categories

3 years ago

A car is traveling 80 km/h if the car's speed remains constant how far will it have traveled in 3 hours

Physics

1 answer:

12345 [234]3 years ago

7 0

C. 80 because it is remaining conctant

You might be interested in

Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor

fenix001 [56]

Answer:

a) $F_{r}= -583.72MN i + 183.47MN j + 6.05GN k$

b) $E=3.04 \frac{GN}{C}$

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

$r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}$

which yields:

$r_{AP}^{2}= 43 m^{2}$

(I will assume the positions are in meters)

Next, we can make use of the force formula:

$F=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

so we substitute the values:

$F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}$

which yields:

$F_{AP}=418.14 MN$

Now we can find its components:

$F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i$

$F_{APx}=191.30 MNi$

$F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j$

$F_{APy}=191.30MN j$

$F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k$

$F_{APz}=318.83 MN k$

And we can now write them together for the first force, so we get:

$F_{AP}=(191.30i+191.30j+318.83k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}$

which yields:

$r_{BP}^{2}= 33 m^{2}$

Next, we can make use of the force formula:

$F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}$

which yields:

$F_{BP}=2.18 GN$

Now we can find its components:

$F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i$

$F_{BPx}=758.98 MNi$

$F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j$

$F_{BPy}=758.98MN j$

$F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k$

$F_{BPz}=1.897 GN k$

And we can now write them together for the second, so we get:

$F_{BP}=(758.98i + 758.98j + 1897k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}$

which yields:

$r_{CP}^{2}= 30 m^{2}$

Next, we can make use of the force formula:

$F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}$

which yields:

$F_{CP}=4.20 GN$

Now we can find its components:

$F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i$

$F_{CPx}=-1.534 GNi$

$F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j$

$F_{CPy}=-766.81 MN j$

$F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k$

$F_{CPz}=3.83 GN k$

And we can now write them together for the third force, so we get:

$F_{CP}=(-1.534i - 0.76681j +3.83k)GN$

So in order to find the resultant force, we need to add the forces together:

$F_{r}=F_{AP}+F_{BP}+F_{CP}$

so we get:

$F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN$

So when adding the problem together we get that:

$F_{r}=(-0.583.72i + 0.18347j +6.05k)GN$

which is the answer to part a), now let's take a look at part b).

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

$F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}$

which yields:

$F_{r}=6.08 GN$

and now we take the formula for the electric field which is:

$E=\frac{F_{r}}{q}$

so we go ahead and substitute:

$E=\frac{6.08GN}{2C}$

$E=3.04\frac{GN}{C}$

7 0

3 years ago

A mass of 0.54 kg attached to a vertical spring stretches the spring 36 cm from its original equilibrium position. The accelerat

UNO [17]

<h2>Spring constant is 14.72 N/m</h2>

Explanation:

We have for a spring

Force = Spring constant x Elongation

F = kx

Here force is weight of mass

F = W = mg = 0.54 x 9.81 = 5.3 N

Elongation, x = 36 cm = 0.36 m

Substituting

F = kx

5.3 = k x 0.36

k = 14.72 N/m

Spring constant is 14.72 N/m

6 0

2 years ago

What is the speed of a transverse wave in a rope of length 3. 1 m and mass 86 g under a tension of 380 n?

Nitella [24]

117 m/sec is the speed of a transverse wave in a rope of length 3. 1 m and mass 86 g under a tension of 380 n.

The wave speed v is given by

v= √τ/μ

where τ is the tension in the rope and μ is the linear mass density of the rope.

The linear mass density is the mass per unit length of rope :

μ= m / L = (0.086 kg)/(3.1 m)=0.0277 kg/m.

v= $\sqrt{ \frac{380 N}{0.0277 kg/m}}$ = 117.125 m/sec (approx. 117 m/sec

In physics, a transverse wave is a wave whose oscillations are perpendicular to the direction of the wave's advance. This is in contrast to a longitudinal wave which travels in the direction of its oscillations. Water waves are an example of transverse wave.

Transverse waves commonly occur in elastic solids due to the shear stress generated; the oscillations in this case are the displacement of the solid particles away from their relaxed position, in directions perpendicular to the propagation of the wave. These displacements correspond to a local shear deformation of the material. Hence a transverse wave of this nature is called a shear wave. Since fluids cannot resist shear forces while at rest, propagation of transverse waves inside the bulk of fluids is not possible.

Learn more about Transverse waves here : brainly.com/question/13761336

#SPJ4

3 0

2 years ago

The minimum amount of energy that has to be added to start a reaction is the ______________ energy.

frosja888 [35]

The best and most correct answer among the choices provided by your question is the first choice or letter A.

<span>The minimum amount of energy that has to be added to start a reaction is the activation energy.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

7 0

3 years ago

Can a object have have zero velocity and nonzero acceleration?

vladimir1956 [14]

Answer:

yes

Explanation:

3 0

3 years ago