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Ber [7]
3 years ago
6

An open container of oil rests on the flatbed of a truck that is traveling along a horizontal road at 55 mi/hr. As the truck slo

ws uniformly to a complete stop in 5 s, what will be the slope of the oil surface during the period of constant deceleration

Physics
1 answer:
rjkz [21]3 years ago
3 0

Answer:

Slope = 0.50.

Explanation:

Below is an attachment containing the solution to the question.

You might be interested in
1 point
marta [7]

Answer:

Let, R be the resistance of the heater wire. Since, two heater wires are of equal length their resistance is also same.

Hence, for series combination of resistances,

R

s

=2R

And for parallel combination of resistances,

R

p

=

2

R

Now, heat produced when they first connected in series is

H

s

=

R

s

V

2

where, V is voltage supplied to the heater.

H

s

=

2R

V

2

.....................................(1)

heat produced when they first connected in series is

H

p

=

R

p

V

2

H

p

=

2

R

V

2

H

p

=

R

2V

2

............................................(2)

From (1) and (2), we get

H

p

H

s

=

R

2V

2

2R

V

2

⇒

H

p

H

s

=

4

1

7 0
3 years ago
Which of the following values represents an index of refraction of an actual material?
serg [7]

Sadly, we're forced to answer the question without the benefit of the
list of choices, which, for some reason, you decided not to let us see.

Index of refraction of a substance =

               (speed of light in vacuum) / (speed of light in the substance).

Any number greater than  ' 1 ' can be an index of refraction.  A number
less than ' 1 ' can't be . . . that would be saying that the speed of light
in this substance is greater than the speed of light in vacuum.

3 0
4 years ago
Find the magnitude of vector A = i - 2j + 3k O V14 10 O4
koban [17]

Answer:

The magnitude of the given vector is \sqrt{14}

Explanation:

For any given vector \overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}

The magnitude is given by

|r|=\sqrt{x^2+y^2+z^2}

comparing with the given vector \overrightarrow{A}=1\widehat{i}-2\widehat{j}+3\widehat{k}

we get

x = 1, y = -2, z = 3

Thus the magnitude of vector A is

|r|=\sqrt{1^2+(-2)^2+3^2}\\\\\therefore |r|=\sqrt{14}

7 0
3 years ago
How do you figure out cos and sin?
dexar [7]

Answer:this is all i can give you

Explanation:

1.The sine of the angle = the length of the opposite side. the length of the hypotenuse.

2.The cosine of the angle = the length of the adjacent side. the length of the hypotenuse.

3.The tangent of the angle = the length of the opposite side. the length of the adjacent side.

5 0
3 years ago
Read 2 more answers
A charge of −25µC is distributed uniformly over the surface of a spherical conductor of radius 12.0 cm. Determine the electric f
Bond [772]

Answer:

a) 0 b) 0 c) -9*10⁵ N/C

Explanation:

a)

  • In electrostatic conditions, no electric field can exist inside a conductor.
  • As the distance r=5 cm falls inside the conductor, the electric field is just zero.

b)

  • Same as above, as r=10 cm is still inside the spherical conductor.

c)

  • At r= 50 cm. from the center of the spherical conductor, we can apply Gauss' Law in order to get the value of the electric field.
  • By symmetry, the electric field, at a same distance from the center, must be radial, and constant on a spherical surface concentric with the spherical conductor.
  • So, we can write the following equation for Gauss'Law:

       \int\ {E} \, dA = \frac{Q}{\epsilon_{0} }

  • If E is constant, we can take it out of the integral, and integrate all the closed spherical surface, as follows:

       E* 4*\pi *r^{2}  =\frac{Q}{\epsilon_{0}}

  • So, we can solve for E, as follows:

       E = \frac{Q}{4*\pi*r^{2}*\epsilon_{0}} =  \frac{(-25e-6)C}{4*\pi*(0.5m)^{2}*8.85e-12C2/N*m2}}\\\\   E = -9e5 N/C

  • E = -9*10⁵ N/C (radially inward, taking the outward direction as positive)
8 0
3 years ago
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