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3 years ago

A car is moving with a speed of 15

1 answer:

5 0

The unit for the speed was not given

If speed is 15km/h
and distance is 1.2km
Then using the formula
Speed = distance/time
Time = t
15 = 1.2/t
15t = 1.2
t = 1.2/15
t = 0.08hrs

But
1hr = 60seconds
0.08hrs = t
t = 60(0.08)
t = 4.8s

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When a potential difference of 154 V is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge d

tester [92]

Answer:

$4.7\mu m$

Explanation:

We are given that

Potential difference=V=154 V

Surface charge density= $\sigma=29nC/m^2=29\times 10^{-5} C/m^2$

Using $1 nC/cm^2=10^{-5} C/m^2$

We know that

$C=\frac{\epsilon_0A}{d}=\frac{Q}{V}=\frac{\sigma A}{V}$

$d=\frac{\epsilon_0V}{\sigma}$

Where

$\epsilon_0=8.85\times 10^{-12}C^2/Nm^2$

Using the formula

$d=\frac{8.85\times 10^{-12}\times 154}{29\times 10^{-5}}$

$d=4.7\times 10^{-6} m=4.7\mu m$

Using $1\mu m=10^{-6} m$

Hence, the spacing between the plates= $4.7\mu m$

3 0

3 years ago

Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 2.88 kg and rotate with

Tanya [424]

Answer:

(a) $K_{small}=4839.3J$

(b) $K_{larger}=17406.4J$

Explanation:

Given data

The angular velocity of two cylinders ω=257 rad/s

The mass of the two cylinders m=2.88 kg

The radius of small cylinder r₁=0.319 m

The radius of larger cylinder r₂=0.605 m

For Part (a)

The rotational kinetic energy of the cylinder is given by:

$K=\frac{1}{2}Iw^2$

Where I is rotational of inertia of solid cylinder about its central axis.

$K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr^2)w^2$

Substitute the given values

$K_{small}=\frac{1}{4}(2.88kg)(0.319)^2(257rad/s)^2 \\K_{small}=4839.3J$

For Part (b)

$K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr_{2}^2)w^2$

Substitute the given values

$K_{larger}=\frac{1}{4}mr_{2}^2w^2\\ K_{larger}=\frac{1}{4}(2.88kg)(0.605m)^2(257rad/s)^2\\ K_{larger}=17406.4J$

8 0

3 years ago

A potato is shot out of cylinder at an angle of 17 degrees above the horizontal with an initial speed of 20 m/s. What is its max

madreJ [45]

Answer:

Maximum height, h = 1.74 meters

Explanation:

It is given that,

A potato is shot out of the cylinder. It is a case of projectile motion. The potato makes an angle of 17 degrees above the horizontal.

Initial speed with which the potato is shot out, u = 20 m/s

We have to find the maximum height of the potato. The maximum height of a projectile (h) is given by the following formula as :

$h=\dfrac{u^2sin^2\theta}{2g}$

Where

$\theta$ = angle between the projectile and the surface

g = acceleration due to gravity

$h=\dfrac{(20\ m/s)^2sin^2(17)}{2\times 9.8\ m/s^2}$

h = 1.74 m

or h = 1.74 meters

Hence, this is the required solution.

8 0

3 years ago

Which of the following describes resistance?

enot [183]

D is your answer hope this helps

8 0

4 years ago

Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug

Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

$F_D=\dfrac{1}{2}\rho C_DA v^2$

$C_D$ =Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

Now by putting the values

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

$\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}$

$\dfrac{F_D_1}{F_D_2}=0.444$

Percentage Change in the drag force

$\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100$

$\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100$

$\Delta F=\dfrac{1-0.444}{0.444}\times 100$

ΔF=125.22 %

Therefore force will increase by 125.22 %.

3 0

3 years ago