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3 years ago

Two 2.5 kg bowling balls are 0.50 m apart. What is the force of gravity of the first bowling ball due to the

Physics

1 answer:

Makovka662 [10]3 years ago

5 0

Answer:The answer is A

Explanation:

To determine the force, we use the equation

Fg= G m1 m2/r2

, where Fg is the force of gravity, G is the gravitational constant,

is the mass of the first body,

is the mass of the second body, and r is the radius. Hence, we have

Fg=6.67×10−11 N×m2/kg2×2.5 kg×2.5 kg/(0.50 m)2=1.7×10−9N

. Since this is a gravitational problem, the force on one body has the same force as the other body, but the directions are opposite.

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The strong nuclear force is ___

Setler [38]

The answer is A) More than 100 times greater than the electrical force

7 0

3 years ago

Read 2 more answers

please write the definition of instantaneous speed, velocity and acceleration and solve problems on them please show workings

Katyanochek1 [597]

Answer:

instantaneous speed: The instantaneous speed is the speed of an object at a particular moment in time. And if you include the direction with that speed, you get the instantaneous velocity. In other words, eight meters per second to the right was the instantaneously velocity of this person at that particular moment in time.

velocity: the speed of something in a given direction.

acceleration: a vehicle's capacity to gain speed within a short time.

solve problems: Problem solving consists of using generic or ad hoc methods in an orderly manner to find solutions to problems.

Explanation:

7 0

3 years ago

A violin with string length 32 cm and string density 1.5 g/cm resonates in its fundamental with the first overtone of a 2.0-m or

love history [14]

Answer:

T=1022.42 N

Explanation:

Given that

l = 32 cm ,μ = 1.5 g/cm

L =2 m ,V= 344 m/s

The pipe is closed so n= 3 ,for first over tone

$f=\dfrac{nV}{4L}$

$f=\dfrac{3\times 344}{4\times 2}$

f= 129 Hz

The tension in the string given as

T = f²(4l²) μ

Now by putting the values

T = f²(4l²) μ

T = 129² x (4 x 0.32²) x 1.5 x 10⁻³ x 100

T=1022.42 N

6 0

3 years ago

One of the solid reactants was treated in a coffee grinder before adding to

Nikitich [7]

Answer:

C. Surface area

Explanation:

The rate of chemical reaction depends on various factors such as:

concentration and pressure
nature of reactants
temperature
surface area
presence of catalyst, etc.

Effect of surface area of reactants: the rate of a chemical reaction can be increased by increasing the the area of contact of the reacting substances. This is especially important when one or more of the reactants are solids., because only the particles of the solid that are exposed are able to take part in the reaction at each instant of time. Therefore, the greater the surface area of the solid reactant particles the faster the reaction.

The surface area of solid reactants can be increased by grinding or pelletizing, thus allowing for a greater contact between the reacting particles,

The instance in which one of the solid reactants was treated in a coffee grinder before adding to the reaction container is one way of increasing the surface area of a reactant.

7 0

3 years ago

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie

IgorLugansk [536]

Answer:

The distance the lighter fragments slides is 343 m

Explanation:

Here, we have

Let the mass of the heavier fragments be m₁

Let the distance the heavy object slide be d₁

Let the mass of the lighter fragments be m₂

Let the final velocity of the heavier fragments be v₁

Let the final velocity of the lighter fragments be v₂

Let the distance the light object slide be d₂

The distance traveled by the heavy fragment = 7.00 m

Therefore since m₁ = 7 × m₂ we have

Initial total momentum = final total momentum

Since the initial total momentum = 0 we have

m₁·v₁ + m₂·v₂ = 0 or

7·m₂·v₁ = -m₂·v₂

∴ v₂ = 7·v₁

The net work done by the heavier block is

W $_{net, 7m}$ = μk × m₁ × g × d₁ = 1/2×m₂×v₂²

Also v₁² = 2μk·g·d₁ and

v₂² = 2μk·g·d₂ so that since v₂ = 7·v₁ or v₁ = v₂/7 we get

(v₂/7)² = 2μk·g·d₁

So v₂²/49 = 2μk·g·d₁

or (2μk·g·d₂)/49 = 2μk·g·d₁

∴ d₂/49 = d₁

d₂ = d₁×49 = 7 × 49 = 343 m

The distance the lighter fragments slides = 343 m.

7 0

3 years ago