Pls someone help 10points!!!

A. Does this graph represent an endothermic or an exothermic reaction? Explain your answer.

ra1l [238]

Answer:

Exothermic reaction

Explanation:

Exothermic reaction relases energy while endothermic absorbs energy.

8 0

3 years ago

A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 50.4 m/s and returns the shot with the ball

jonny [76]

(a) The impulse delivered to the ball by the racket is 5.24 kg.m/s

(b) The work that the racket does on the ball is -35.1 Joule

<h3>Further Explanation</h3>

<u>Given :</u>

mass of ball = m = 0.06 kg

initial velocity = v₁ = -50.4 m/s

final velocity = v₂ = 37.0 m/s

<u>Unknown :</u>

(a) Impulse = I = ?

(b) Work = W = ?

<u>Solution :</u>

<h2>Question (a) :</h2>

In this question , we could use the formula from Second Law of Newton :

$I = \Delta p$

$I = p_2 - p_1$

$I = m \times v_2 - m \times v_1$

$I = m \times (v_2 - v_1)$

$I = 0.06 \times (37.0 - (-50.4))$

$I = 0.06 \times (87.4)$

$I = 5.244~kg.m/s$

$\large { \boxed {I \approx 5.24~kg.m/s} }$

<h2>Question (b) :</h2>

$W = F \times d$

$W = (\frac{I}{\Delta t})(\frac {v_1 + v_2}{2} \Delta t)$

$W = \frac{I(v_1 + v_2)}{2}$

$W = \frac{5.244(-50.4 + 37)}{2}$

$W = \frac{5.244(-13.4)}{2}$

$W = -35.1348~Joule$

$\large { \boxed {W \approx -35.1~Joule}}$

<h3>Learn more</h3>

Newton's Law of Motion: brainly.com/question/10431582

Example of Newton's Law: brainly.com/question/498822

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

Keywords: Newton, Law, Impulse, Work

6 0

4 years ago

Which is a physical property that can be used to tell one substance from another?

siniylev [52]

Should be B. Density.

3 0

3 years ago

Read 2 more answers

The expanding gases that leave the muzzle of a rifle also contribute to the recoil. A .30 caliber bullet has mass 7.20×10−3 kg a

katen-ka-za [31]

Answer:

Mg= 1.3418 kg*m/s

Explanation:

Mass of rifle

$m_{r}=2.90 kg$

Recoil velocity of rifle

$v_{r}=1.95 m/s$

Momentum of rifle

$m*v_{r}=3.0kg*1.95\frac{m}{s}==5.85 \frac{kg*m}{s}$

mass of bullet

mb=7.2x10^{-3} kg=0.0072 kg

velocity of bullet relative to muzzel

v_{b} = 601 m/s

velocity of bullet relative to earth= 601 - 1.95=599.05 m/s

momentum of bullet

$m_{b}*m_{b}=7.2x10^{-3}kg*599.05\frac{m}{s}=4.3132 \frac{kg*m}{s}$

the momentum of the propellant gases = momentum of rifle - momentum of bullet

the momentum of the propellant gases

Mg=5.655 -4.3132

Mg= 1.3418 kg*m/s

the momentum of the propellant gases in a coordinate system attached to the earth as they leave the muzzle of the rifle is 1.3418 kgm/s

5 0

3 years ago

An object whose mass is 300 lb experiences changes in its kinetic and potential energies owing to the action of a resultant forc

aev [14]

Answer: initial speed = 151.89 ft/s

Explanation:

1) About the question:

The question is incomplete. The final missing part is: If the object's elavation increases by 100 ft and its final velocirty is 200 ft/s, what is its initial velocity, in ft/s? Let g= 32.2 ft/s^2

2) Physical principle:

Work = change in mechanical energy

3) Formulas

Kinetic energy, KE = (1/2)mv²

Potential energy, PE = mgh

Mechanical energy, ME = KE + PE

4) Conversion of units:

i) 300 lbs = 300/2.2046 kg = 136.08 kg

ii) 140 btu = 140 × 1055.06 J = 147,708.4 J

iii) g = 9.81 m/s²

iv) 100 ft = 100 × 0.3048m = 30.48 m

v) 200 ft/s = 200 × 0.3048 m/s = 60.96 m/s

5) Solution

i) change in PE:

ΔPE = mgΔh = 136.08kg × 9.81 m/s² × 30.48m = 40,689.12 J

ii) change in KE

ΔKE = work done on the object - ΔPE = 147,708.4J - 40,689.12J = 107,019.28 J

iii) initial KE

initial KE = final KE - ΔKE = (1/2)mv² - ΔKE = (1/2) (136.08kg)(60.96m/s)² = 252,844.91J - 107,019.28J = 145,825.63J

iv) initial speed

KE = (1/2)mv² ⇒ v² = 2KE / m = 2 (145,825.63J) / 136.08 kg = 2,143.23 m²/s²

v = √v² = 46.30 m/s

v) Convert to ft/s

46.30 m/s = 46.30 / 0.3048 ft/s = 151.89 ft/s

7 0

4 years ago