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3 years ago

Which particle in the atom has the smallest mass? electron proton neutron photon

Physics

2 answers:

Ray Of Light [21]3 years ago

6 0

Answer: It would be A: electron

I hope this helps!

HACTEHA [7]3 years ago

3 0

Answer:

Photon

Explanation:

Electrons don't weigh very much at all, and don't account for the chemical mass of an element. However, Photons don't have ANY mass.

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A hypothesis is a(n) ______________

marusya05 [52]

Answer:

It is an educated guess

Explanation:

let me know if the bottom ones need to be answered too.

6 0

3 years ago

Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an

joja [24]

Answer:

Approximately $9.7\; \rm m \cdot s^{-2}$ .

Explanation:

Assuming that there is no other force on this vehicle, the $16000\; \rm N$ force from the road would be the only force on this vehicle. The net force would then be equal to this $16000\; \rm N\!$ force. The size of the net force would be $16000\; \rm N\!\!$ .

Let $m$ denote the mass of this vehicle and let $\Sigma F$ denote the net force on this vehicle.

By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, $a$ , would be:

$\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}$ .

For this vehicle, $\Sigma F = 16000\; \rm N$ whereas $m = 1650\; \rm kg$ . The acceleration of this vehicle would be:

$\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}$ .

8 0

3 years ago

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.00 m/s, and

Leya [2.2K]

Answer:

A) 0.9844 s

B) x2 = 0.4587 m

C) v = 6.657 m/s

Explanation:

We are given;

Height of take off point above pool; x1 = 1.8 m

Initial take off velocity; u = 3 m/s

Final velocity at highest point before free fall; v = 0 m/s

B) To find the highest point above the board her feet reaches means the distance from take off to the top of the motion just before free fall.

Thus, we will be using equation of motion and we have;

v² = u² + 2gs

Now, let s = x2 which will be the distance between take off and the top before free fall.

So;

v² = u² + 2g(x2)

Now,since the motion is against gravity, g will be negative.

Thus;

v² = u² + 2(-9.81)(x2)

Plugging in the relevant values to give;

0² = 3² - (19.62x2)

19.62(x2) = 9

x2 = 9/19.62

x2 = 0.4587 m

A) We want to find how long her feet is in air.. It means we want to find out the time to get to a distance of x1 and also the time to achieve the distance (x1 + x2) on free-fall.

Thus, using equation of motion;

v = u + gt

Again, g = -9.81

Thus;

0 = 3 - 9.81t1

9.81t1 = 3

t1 = 3/9.81

t1 = 0.3058 s

Now, for the time taken to achieve the distance (x1 + x2) on free-fall, we will use the formula;

s = ut + ½gt²

Where s = (x1 + x2) = 1.8 + 0.4587 = 2.2587 m

And now, u = 0 m/s because the start of the free fall is from maximum height with velocity of 0 m/s. Again, g = - 9.81 m/s²

Thus;

2.2587 = 0 - ½(-9.81)(t2)²

2.2587 = 4.905(t2)²

(t2)² = 2.2587/4.905

(t2)² = 0.4605

t2 = √0.4605

t2 = 0.6786 s

Thus, total time of feet in air = t1 + t2 = 0.3058 + 0.6786 = 0.9844 s

C) Velocity when feet hit the water would be given by;

v = u + gt

Where u = 0 m/s and t = t2 = 0.6786

Since it's in direction of gravity, g = 9.81 m/s

v = 0 + (0.6786 × 9.81)

v = 6.657 m/s

4 0

3 years ago

Will name brainliest please pleas answer

Greeley [361]

3 is the answer teeeeeeeeeheeeeeeeee

6 0

3 years ago

The International Space Station (ISS) orbits Earth at an altitude of 4.08 × 105 m above the surface of the planet. At what veloc

alukav5142 [94]

This question involves the concepts of orbital velocity and orbital radius.

The orbital velocity of ISS must be "7660.25 m/s".

The orbital velocity of the ISS can be given by the following formula:

$v=\sqrt{\frac{GM}{R}}$

where,

v = orbital velocity = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of Earth = 5.97 x 10²⁴ kg

R = orbital radius = radius of earth + altitude = 63.78 x 10⁵ m + 4.08 x 10⁵ m

R = 67.86 x 10⁵ m

Therefore,

$v=\sqrt{\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}{67.86\ x\ 10^5\ m}}$

Learn more about orbital velocity here:

brainly.com/question/541239

3 0

2 years ago