Is the following sentence true or false? Two atoms of the same element can have different numbers of protons.

Question 9

Anastasy [175]

nAnswer:

uyui

Explanation:

4 0

3 years ago

A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend.

MatroZZZ [7]

Answer: Part(a)=0.041 secs, Part(b)=0.041 secs

Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction

now we know that

a=-9.81 m/s^2 ( negative because it is pulling the player downwards)

we also know that

s=76 cm= 0.76 m ( maximum s)

using kinetic equation

$v^2=u^2+2as$

where v is final velocity which is zero at max height and u is it initial

hence

$u^2=-2(-9.81)*0.76$

$u=3.8615 m/s\\$

now we can find time in the 15 cm ascent

$s=ut+0.5at^2$

$0.15=3.861*t+0.5*9.81t^2\\$

using quadratic formula

$t=\frac{-3.861+\sqrt{3.86^2-4*0.5*9.81(-0.15)} }{2*0.5*9.81}$

t=0.0409 sec

the answer for the part b will be the same

To find the answer for the part b we can find the velocity at 15 cm height similarly using

$v^2=u^2+2as$

where s=0.76-0.15

as the player has traveled the above distance to reach 15cm to the bottom

$v^2=0^2 +2*(9.81)*(0.76-0.15)$

$v=3.4595$

when the player reaches the bottom it has the same velocity with which it started which is 3.861

hence the time required to reach the bottom 15cm is

$t=\frac{3.861-3.4595}{9.81}$

t=0.0409

8 0

3 years ago

Z. A force that gives a 8-kg objet an acceleration of 1.6 m/s^2 would give a 2-kg object an

Paladinen [302]

Answer:

$\boxed {\boxed {\sf D.\ 6.4\ m/s^2}}$

Explanation:

We need to find the acceleration of the 2 kilogram object. Let's complete this in 2 steps.

<h3>1. Force of 1st Object </h3>

First, we can find the force of the first, 8 kilogram object.

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

$F=m \times a$

The mass of the object is 8 kilograms and the acceleration is 1.6 meters per square second.

m= 8 kg
a= 1.6 m/s²

Substitute these values into the formula.

$F= 8 \ kg * 1.6 \ m/s^2$

Multiply.

$F= 12.8 \ kg*m/s^2$

<h3>2. Acceleration of the 2nd Object </h3>

Now, use the force we just calculated to complete the second part of the problem. We use the same formula:

$F= m \times a$

This time, we know the force is 12.8 kilograms meters per square second and the mass is 2 kilograms.

F= 12.8 kg *m/s²
m= 2 kg

Substitute the values into the formula.

$12.8 \ kg*m/s^2= 2 \ kg *a$

Since we are solving for the acceleration, we must isolate the variable (a). It is being multiplied by 2 kg. The inverse of multiplication is division. Divide both sides of the equation by 2 kg.

$\frac {12.8 \ kg*m/s^2}{2 \ kg}= \frac{2\ kg* a}{2 \ kg}$

$\frac {12.8 \ kg*m/s^2}{2 \ kg}=a$

The units of kilograms cancel.

$\frac {12.8}{2}\ m/s^2=a$

$6.4 \ m/s^2=a$

The acceleration is 6.4 meters per square second.

4 0

3 years ago

radio waves are electromagnetic waves that travel at the speed of light 300 000 kilometers per second what is the wave length of

Anarel [89]

<h2>Answer:</h2>

<h3>The wavelength of 100-MHz radio waves is 3 m, yet using the sensitivity of the resonant frequency to the magnetic field strength, details smaller than a millimeter can be imaged.</h3>

<h2>Hope this helps you ❤️</h2>

<h2>MaRk mE aS braiNliest ❤️</h2>

3 0

3 years ago

You toss a 0.40-kg ball at 9.0 ms/ to a 14-kg dog standing on an iced-over pond. The dog catches the ball and begins to slide on

just olya [345]

Answer:

a) $v_{f}$ = 0.25 m / s b) u = 0.25 m / s

Explanation:

a) To solve this problem let's start with the conservation of the moment, for this we define a system formed by the ball plus the dog, in this case all the forces are internal and the moment is conserved

We will write the data

m₁ = 0.40 kg

v₁₀ = 9.0 m / s

m₂ = 14 kg

v₂₀ = 0

Initial

po = m₁ v₁₀

Final

$p_{f}$ = (m₁ + m₂) vf

po = pf

m₁ v₁₀ = (m₁ + m₂) $v_{f}$

$v_{f}$ = v₁₀ m₁ / (m₁ + m₂)

$v_{f}$ = 9.0 (0.40 / (0.40 +14)

$v_{f}$ = 0.25 m / s

b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass

u = 0.25 m / s

In the direction of movement of the ball

c) Let's calculate the kinetic energy in both moments

Initial

K₀ = ½ m₁ v₁₀² +0

K₀ = ½ 0.40 9 2

K₀ = 16.2 J

Final

$K_{f}$ = ½ (m₁ + m₂) $v_{f}$ 2

$K_{f}$ = ½ (0.4 +14) 0.25 2

$K_{f}$ = 0.45 J

ΔK = K₀ - $K_{f}$

ΔK = 16.2-0.445

ΔK = 1575 J

These will transform internal system energy

d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.

v₁₀’= v₁₀ -u

v₁₀’= 9 -.025

v₁₀‘= 8.75 m / s

v₂₀ ‘= v₂₀ -u

v₂₀‘= - 0.25 m / s

$v_{f}$ ‘= $v_{f}$ - u

$v_{f}$ = 0

Initial

K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²

Ko = ½ 0.4 8.75² + ½ 14.0 0.25²

Ko = 15.31 + 0.4375

K o = 15.75 J

Final

$k_{f}$ = ½ (m₁ + m₂) vf’²

$k_{f}$ = 0

All initial kinetic energy is transformed into internal energy in this reference system

3 0

3 years ago