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4 years ago

If the distance between two objects is increased,the gravitational attraction between them will.....?

Physics

2 answers:

Archy [21]4 years ago

5 0

Attraction will decrease

tia_tia [17]4 years ago

4 0

The attraction will decrease. Hope I helped :)

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A banana has a mass of 1.4 kg. What is its weight, in Newtons?

Alenkasestr [34]

Answer

given :

m=1.4 kg

r/q = N

N = m(9.8)

= 1.4 kg(9.8)

= 13.72 Newton

7 0

3 years ago

Read 2 more answers

An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displace

Zepler [3.9K]

Answer:

Explanation:

Displacement is the shortest distance possible between initial and final position.

As the athlete is in circular motion , displacement is zero. [ THis is because , the initial and final position is the same ]

Distance covered in 1 round = Circumference of the circle

Circumference of the circle = 2πr

Diameter = 200 m

Radius = 100 m

Distance covered in 1 round[ 40 sec ] = 2 × 22/7 × 100 = 628.57 m

Distance covered in 1 sec = 628.57 ÷ 40 = 15.71 m

2min 20 sec = 140 sec

Distance covered in 140 sec = 15.71 × 140 = 2199.4 m

For each complete round the displacement is zero. Therefore for 3 complete rounds, the displacement will be zero.

At the end of his motion, the athlete will be in the diametrically opposite position. That is, displacement = diameter = 200 m.

Hence, the distance covered is 2200 m and the displacement is 200 m.

5 0

4 years ago

When a 4.32 kg object is hung vertically on a certain light spring that obeys Hooke's Law, the spring stretches 2.92 cm.

Alika [10]

(a) 1.01 cm

First of all, we need to find the spring constant of the spring.

The force initially applied to the spring is equal to the weight of the block hanging on it:

$F=mg=(4.32)(9.8) = 42.3 N$

where m = 4.32 kg is the mass of the block and g = 9.8 m/s^2 is the acceleration of gravity.

When this force is applied, the spring stretches by

$x=2.92 cm = 0.0292 m$

We can find the spring constant by using Hooke's law:

$F=kx$

where k is the spring constant. Solving for k,

$k=\frac{F}{x}=\frac{42.3}{0.0292}=1448.6 N/m$

Later, the first object is removed and another object of mass

m' = 1.50 kg

is hung on the spring. The weight of this object is

$F'=m'g=(1.50)(9.8)=14.7 N$

So, if we use Hooke's law again, we can find the new stretching of the spring:

$x'=\frac{F'}{k}=\frac{14.7}{1448.6}=0.0101 m = 1.01 cm$

(b) 1.16 J

The work that must be done on the spring is equal to the elastic potential energy that would be stored in the spring, therefore:

$W=\frac{1}{2}kx^2$

where we have

k = 1448.6 N/m is the spring constant

x = 4.00 cm = 0.04 m is the new stretching

Solving the equation, we find the work that must be done by the external force:

$W=\frac{1}{2}(1448.6)(0.04)^2=1.16 J$

6 0

3 years ago

Dock walls sometimes have pillars replaced with rubber car tyres.explain how this reduces the possibility of damage what a boat

ra1l [238]

The rubber absorbs the kinetic energy from the impact of a boat hitting the tires

8 0

3 years ago

If there's air in a ball and it bounces how come a room doesn't bounce

finlep [7]

Answer:

Funny answer but works!

Explanation:

Not enough air. Go outside, crack open the window and blow air into the room like you would a balloon then close the window quickly. If it doesn't start bouncing, try giving the house a push. Good luck.

4 0

3 years ago