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SOVA2 [1]
3 years ago
13

27. The electric field around a positive charge is shown in the diagram. Describe the nature of these lines. Please use 2 conten

t related sentences

Physics
1 answer:
KIM [24]3 years ago
4 0

Answer:

Once we place a positive test at a point close to the sphere, we find that an electrostatic force is applied to the outside of the sphere. Therefore, at any point around the sphere, the electric field vector is radially outward.

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A bowling ball is launched from the top of a building at an angle of 35° above the horizontal with an initial speed of 15 m/s. T
Mamont248 [21]

Let y_0 be the height of the building and thus the initial height of the ball. The ball's altitude at time t is given by

y=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ\,t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity.

The ball reaches the ground when y=0 after t=2.9\,\mathrm s. Solve for y_0:

0=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ(2.9\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(2.9\,\mathrm s)^2

\implies y_0\approx16.258\,\mathrm m

so the building is about 16 m tall (keeping track of significant digits).

3 0
2 years ago
(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th
julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

V = -\dfrac{kQ}{r}

where k = 9\times 10^9 \text{ F/m}

Difference in potential between the points is

kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}

PD = 135\times 10^3\text{ V} = 135\text{ kV}

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]

10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10

\dfrac{1}{x} = 0

x = \infty

The charge chould be moved to infinity

7 0
3 years ago
Two forces act on an object. One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N direct
xeze [42]

Answer:

8 N North.

Explanation:

Given that,

One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N directed south.

We need to find the magnitude of net force acting on the object.

Let North is positive and South is negative.

Net force,

F = 10 N +(-2 N)

= 8 N

So, the magnitude of net force on the object is 8 N and it is in North direction (as it is positive). Hence, the correct option is (d) "8N north".

7 0
2 years ago
A heat engine does 9200 J of work per cycle while absorbing 22.0 kcal of heat from a hightemperature reservoir. What is the effi
Morgarella [4.7K]

Answer: 9.9%

Explanation: efficiency = (work output /work input) × 100

Note that, 1 kilocalorie = 4184 joules, hence 22kcal = 22× 4184 = 92048 joules.

Work output = 9200 j and work input = 92048 j

Efficiency = (9200/92048) × 100 = 0.099 × 100 = 9.9%

4 0
3 years ago
Suppose a car of mass m is moving at a constant speed v of
SIZIF [17.4K]

Answer:

The angle of banked curve that makes the reliance on friction unnecessary is

\arcsin(v^2/(gR))

Explanation:

In order the car to stay on the curve without friction, the net force in the direction of radius should be equal or smaller than the centripetal force. Otherwise the car could slide off the curve.

The only force in the direction of radius is the sine component of the weight of the car

w_r = mg\sin(\theta)

The cosine component is equivalent to the normal force, which we will not be using since friction is unnecessary.

Newton’s Second Law states that

F_{net} = ma = mg\sin(\theta)\\\sin(\theta) = a/g

Also, the car is making a circular motion:

a = \frac{v^2}{R}

Combining the equations:

\sin(\theta) = \frac{a}{g} = \frac{v^2/R}{g} = \frac{v^2}{gR}

Finally the angle is

\arcsin(v^2/(gR))

4 0
3 years ago
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