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3 years ago

6

What would happen if a proton were added to the nucleus of the atom? (Check all that apply)

1 answer:

katen-ka-za [31]3 years ago

3 0

Answer:

It would become a different element.

Explanation:

When you change the number of protons i.e, add a proton in the nucleus of an atom, you will change the atom from one element to a different element and also positive charge of the atom also increased by 1.

For example, adding a proton to the nucleus of an atom of hydrogen creates an atom of helium.

So, we conclude that by adding a proton the atom will change from one element to another element.

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Two bicycle tires are set rolling with the same initial speed of 3.30 m/s along a long, straight road, and the distance each tra

vredina [299]

Answer:

At low pressure- $\mu_{k}=0.02315$

At high pressure- $\mu_{k}=0.00445$

Explanation:

Initial speed, $V_{i}=3.3 m/s$

Final speed, $V_{f}=3.3/2= 1.65 m/s$

Net horizontal force due to rolling friction $F_{net}=\mu_{k}$ mg where m is mass, g is acceleration due to gravity, $\mu_{k}$ is coefficient of rolling friction

From kinematic relation, $V_{f}^{2}- V_{i}^{2}=2ad$

For each tire,

$V_{f}^{2}- V_{i}^{2}=2\mu_{k}gd$

Making $\mu_{k}$ the subject

$\mu_{k}=\frac {V_{f}^{2}- V_{i}^{2}}{2gd}$

Under low pressure of 40 Psi, d=18 m

$\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*18}=-0.02315$

Therefore, $\mu_{k}=0.02315$

At a pressure of 105 Psi, d=93.7

$\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*93.7}=-0.00445$

Therefore, $\mu_{k}=0.00445$

4 0

3 years ago

Cooling down after a workout allows the oxygen to continuing moving through the body and keeps the muscles from tightening up to

Liono4ka [1.6K]

Answer: true

Explanation:

7 0

3 years ago

Read 2 more answers

A mobile phone is pulled northward by a force of 10 n and at the same time pulled southward by another force of 15 n. the result

makvit [3.9K]

5 Southward would be the correct answer.

6 0

3 years ago

What is the approximate value of the gravitational force between a 73 kg astronaut and a 7.1×104 kg spacecraft when they're 89 m

Luden [163]

Answer:

$F = 4.3671 * 10^{-8}\ Newtons$

Explanation:

The gravitational force between two corpses is given by the following equation:

$F = GMm/d^2$

Where F is the force, G is the gravitational constant

( $G = 6.67408*10^{-11}\ m^3kg^{-1}s^{-2}$ ), M and m are the masses of the corpses and d is the distance between them.

So we have that:

$F = 6.67408*10^{-11} * 7.1*10^4 * 73/89^2$

$F = 4.3671 * 10^{-8}\ Newtons$

5 0

3 years ago

A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long a

IrinaK [193]

The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

To find the answer, we need to know about the time of flight and range of projectile motion.

<h3>What's the expression of range of a projectile motion?</h3>

Range = U²× sin(2θ)/g
U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
U=√{Range×g/sin(2θ)}
Here, range= 2.20m, = 36.5°
U= √{2.20×9.8/sin(73)}

U= √{2.20×9.8/sin(73)} = 22.5m/s

<h3>What's the expression of time of flight in projectile motion?</h3>

Time of flight= (2×U×sinθ)/g
So, T= (2×22.5×sin36.5°)/9.8

= 2.73 s

Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

Learn more about the range and time period of projectile motion here:

brainly.com/question/24136952

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4 0

2 years ago