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2 years ago

What does the half life for a radioactive material mean?

1 answer:

6 0

Half-life, in radioactivity, the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay (change spontaneously into other nuclear species by emitting particles and energy), or, equivalently, the time interval required for the number of disintegrations per second of a radioactive

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3. A worker pushes a box across the floor to the right at a constant speed with a force of 25N. What

WITCHER [35]

Answer:

Friction between the box and the floor is 25N to the left.

Explanation:

According to Newton's second law of motion, the net force acting on an object is equal to the produce between the object's mass and its acceleration:

$F_{net}=ma$

where

m is the mass of the object

a is its acceleration

In this problem, we have two forces acting on the object:

- The applied force, F = 25 N, to the right

- The force of friction $F_f$ , opposing the motion of the box, so to the left

So we can write the net force as

$F_{net}=F-F_f$

Also, we know that the box is moving at constant speed: this means its acceleration is zero, so

$a=0$

Therefore

$F_{net}=0$

WHich means:

$F-F_f=0$

And therefore,

$F_f=F=25 N$

which means that the force of friction is also 25 N.

6 0

2 years ago

Please help me with this please <br><br> I’ll mark you Brainly

ivolga24 [154]

I think it is option (C).

If the answer is helpful then mark me as brainly.

4 0

2 years ago

A bowling ball that has a radius of 11.0 cm and a mass of 5.00 kg rolls without slipping on a level lane at 2.80 rad/s.

NemiM [27]

Answer:

$\dfrac{K_t}{K_r}=\dfrac{5}{2}$

Explanation:

Given that,

Mass of the bowling ball, m = 5 kg

Radius of the ball, r = 11 cm = 0.11 m

Angular velocity with which the ball rolls, $\omega=2.8\ rad/s$

To find,

The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.

Solution,

The translational kinetic energy of the ball is :

$K_t=\dfrac{1}{2}mv^2$

$K_t=\dfrac{1}{2}m(r\omega)^2$

$K_t=\dfrac{1}{2}\times 5\times (0.11\times 2.8)^2$

The rotational kinetic energy of the ball is :

$K_r=\dfrac{1}{2}I \omega^2$

$K_r=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\times \omega^2$

$K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 5\times (0.11)^2\times (2.8)^2$

Ratio of translational to the rotational kinetic energy as :

$\dfrac{K_t}{K_r}=\dfrac{5}{2}$

So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2

4 0

3 years ago

How to calculate the angle of a resultant<br>

stiv31 [10]

Answer:

The direction angle θ of the resultant in the Polar (positive) specification is then θ = α + 60°. The Law of Cosines is used to calculate the magnitude (r) and the Law of Sines is used to calculate the angle (α).

3 0

2 years ago

When sitting in the tree the cat has a total of 1375J in its Gravitational potential store. What is the maximum amount of energy

Murrr4er [49]

Answer:

1375J

Explanation:

The gravitational potential/potential energy of the at the top of the tree which is the energy by virtue of its position.

P.E = mgh

mass = m

Acceleration due to gravity = g

height = h

At the top of the tree, the value of h (height) is high resulting in the gravitational potential. When the cat lands on the ground, the value of h is zero, the the gravitational potential would be zero and all the potential energy have been converted to other forms of energy.

Therefore, the total gravitational potential store is equal to the maximum amount of energy that can be transferred which is equal to 1375J.

4 0

2 years ago