How can electric shock be prevented?

a car moving on a road passes by kilometer 218 at 10:15 am and milestone 236 at 10:30 am. determine an average scalar speed in k

Leno4ka [110]

Answer:

v?

Explanation:

v = ∆s / ∆t

v =?

∆s = 236 - 218 = 18km

=t = 10: 30 - 10:15 = 15 minutes to hour 15/60 = 0.25h

v = 18 / 0.25

v = 72km / h

speed of 72km / h

5 0

3 years ago

Which describes an object in projectile motion? Check all that apply.A.Gravity acts to pull the object downB.The object moves in

irina1246 [14]

Answer:

A.Gravity acts to pull the object down

D.The object’s inertia carries it forward.

E.The path of the object is curved.

Explanation:

The motion of a projectile consists of two independent motions:

- A uniform motion along the horizontal direction, with constant horizontal speed

- A vertical motion with constant acceleration of g = 9.8 m/s^2 downward (acceleration due to gravity), due to the presence of the force of gravity, so the vertical velocity changes (increases in the downward direction)

As a result, the combined motion of the projectile has a curved trajectory (parabolic, more specifically). So the following options are correct:

A.Gravity acts to pull the object down --> gravity acts along the vertical direction

D.The object’s inertia carries it forward. --> there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal motion continues with constant speed

E.The path of the object is curved

3 0

3 years ago

A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains 0.1

Shkiper50 [21]

Answer:

0.10013 atm

Explanation:

Applying Boyle's Law,

P'V' = PV................... Equation 1

Where P' = Initial pressure of air, V' = Initial volume of air, P = Final pressure of air, V = Final volume of air.

make P the subject of the equation

P = P'V'/V..................... Equation 2

Given: P' = 0.355 atm, V' 0.110 m³, V = 0.390 m³

Substitute into equation 2

P = 0.355(0.11)/0.39

P = 0.10013 atm.

5 0

3 years ago

Two charged concentric spherical shells have radii of 11.0 cm and 14.0 cm. The charge on the inner shell is 3.50 ✕ 10−8 C and th

Sergio039 [100]

Answer:

The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

Explanation:

Given that,

Radius of inner shell = 11.0 cm

Radius of outer shell = 14.0 cm

Charge on inner shell $q_{inn}=3.50\times10^{-8}\ C$

Charge on outer shell $q_{out}=1.60\times10^{-8}\ C$

Suppose, at r = 11.5 cm and at r = 20.5 cm

We need to calculate the magnitude of the electric field at r = 11.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Where, q = charge

k = constant

r = distance

Put the value into the formula

$E=\dfrac{9\times10^{9}\times3.50\times10^{-8}}{(11.5\times10^{-2})^2}$

$E=2.38\times10^{4}\ N/C$

The total charge enclosed by a radial distance 20.5 cm

The total charge is

$q=q_{inn}+q_{out}$

Put the value into the formula

$q=3.50\times10^{-8}+1.60\times10^{-8}$

$q=5.1\times10^{-8}\ C$

We need to calculate the magnitude of the electric field at r = 20.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times5.1\times10^{-8}}{(20.5\times10^{-2})^2}$

$E=1.09\times10^{4}\ N/C$

Hence, The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

7 0

3 years ago

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 N , one at an angle 15.0 ∘ west of north,

Darina [25.2K]

Answer:

The total work done by the two tugboats on the supertanker is 3.44 *10^9 J

Explanation:

The force by the tugboats acting on the supertanker is constant and the displacement of the supertanker is along a straight line.

The angle between the 2 forces and displacement is ∅ = 15°.

First we have to calculate the work done by the individual force and then we can calculate the total work.

The work done on a particle by a constant force F during a straight line displacement s is given by following formula:

W = F*s

W = F*s*cos∅

With ∅ = the angles between F and s

The magnitude of the force acting on the supertanker is F of tugboat1 = F of tugboat 2 = F = 2.2 * 10^6 N

The total work done can be calculated as followed:

Wtotal = Ftugboat1 s * cos ∅1 + Ftugboat2 s* cos ∅2

Wtotal = 2Fs*cos∅

Wtotal = 2*2.2*10^6 N * 0.81 *10³ m s *cos15°

Wtotal = 3.44*10^9 Nm = <u>3.44 *10^9 J</u>

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The total work done by the two tugboats on the supertanker is 3.44 *10^9 J

5 0

2 years ago