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2 years ago

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Two charged spheres are 8.45 cm apart. They are moved, and the force on each of them is found to have been tripled. How far apar

t are they now?

1 answer:

Aleonysh [2.5K]2 years ago

4 0

Force between two charges is given by

$F = \frac{kq_1q_2}{r^2}$

here force is inversely depends on the square of the distance

so here if distance is decreased the force will increase'

now if force is tripled

$\frac{3F}{F} = \frac{r^2}{r'^2}$

initially distance between charges was 8.45 cm

$3 = \frac{8.45^2}{r'^2}$

$r' = \frac{8.45}{\sqrt3}$

$r' = 4.88 cm$

so the distance between charges is 4.88 cm now

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Which of the following are dwarf planets? Check all that apply. Ceres Namaka Eris Charon Haumea Makemake Pluto

kicyunya [14]

Ceres: Yes!
Namaka: No!
Eris: Yes!
Charon: No. (it's a satellite, and dwarf planet's can't be satellites!)
Haumea: Yes!
Makemake: Yes!
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1 year ago

Read 2 more answers

In 1993, Ileana Salvador of Italy walked 3.0 km in under 12.0 min. Suppose that during 115 m of her walk Salvador is observed to

dexar [7]

Answer:

t = 25 seconds

Explanation:

Given that,

Distance, d = 115 m

Initial speed, u = 4.2 m/s

Final speed, v = 5 m/s

We need to find the time taken in increasing the speed.

We know that,

Acceleration, $a=\dfrac{v-u}{t}$ ....(1)

The third equation of kinematics is as follows :

$v^2-u^2=2ad\\\\\text{Put the value of a in above equation}\\\\v^2-u^2=2\times \dfrac{v-u}{t}\times d\\\\\because (a^2-b^2)=(a-b)(a+b)\\\\(v-u)(v+u)=\dfrac{2\times (v-u)d}{t}\\\\t=\dfrac{2d}{v+u}\\\\\text{Putting all the values}\\\\t=\dfrac{2\times 115}{4.2+5}\\\\t=25\ s$

Hence, it will take 25 seconds to increase the speed.

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1 year ago

Two children stand on a platform at the top of a curving slide next to a backyard swimming pool. At the same moment the smaller

victus00 [196]

1.

Answer:

a) It is less

Explanation:

By energy conservation we can say that initial potential energy of both child must be equal to the final kinetic energy of the two child.

Since initially they are at same height so we will say that initial potential energy will be given as

$mgH$ and MgH

so the child with greater mass has more energy and hence smaller child will reach with smaller kinetic energy

2.

Answer:

b. The two speeds are equal.

Explanation:

As we know by mechanical energy conservation law we have

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh}$

since both child starts at same height so here they both will reach the bottom at same speed

3.

Answer:

c. The two accelerations are equal

Explanation:

Since we know that average acceleration of the motion is given as

$a = \frac{v_f - v_i}{\Delta t}$

since here initial and final speeds are same so they both must have same average acceleration here.

5 0

2 years ago

A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet

Gnom [1K]

Answer:

d. 37 °C

Explanation:

$m_{m}$ = mass of lump of metal = 250 g

$c_{m}$ = specific heat of lump of metal = 0.25 cal/g°C

$T_{mi}$ = Initial temperature of lump of metal = 70 °C

$m_{w}$ = mass of water = 75 g

$c_{w}$ = specific heat of water = 1 cal/g°C

$T_{wi}$ = Initial temperature of water = 20 °C

$m_{c}$ = mass of calorimeter = 500 g

$c_{c}$ = specific heat of calorimeter = 0.10 cal/g°C

$T_{ci}$ = Initial temperature of calorimeter = 20 °C

$T_{f}$ = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

$m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) + m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C$

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2 years ago

The net force on a boat causes it to accelerate at 1.55 m/s2. The mass of the boat is 215 kg. The same net force causes another

jekas [21]

Answer:

2666 kg

0.11567 m/s²

Explanation:

m = Mass of boat

a = Acceleration of boat

From Newton's second law

Force

$F=ma\\\Rightarrow F=215\times 1.55\\\Rightarrow F=333.25\ N$

Force on the first boat is 333.25 N

$F=ma\\\Rightarrow m=\frac{F}{a}\\\Rightarrow m=\frac{333.25}{0.125}\\\Rightarrow m=2666\ kg$

Hence, mass of the second boat is 2666 kg

Combined mass = 2666+215 = 2881 kg

$F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{333.25}{2881}\\\Rightarrow a=0.11567\ m/s^2$

The acceleration on the combined mass is 0.11567 m/s²

6 0

1 year ago