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soldi70 [24.7K]

3 years ago

11

If a small sports car collides head-on with a massive truck, which vehicle experiences the greater impact force? Which vehicle e

xperiences the greater acceleration? Explain your answers.

1 answer:

garri49 [273]3 years ago

5 0

Answer:

Small sports car.

Explanation:

Lets take

mass of the small car = m

mass of the truck = M

As we know that when car collide with the massive truck then due to change in the moment of the car both car as well as truck will feel force.We also know that from Third law of Newton's ,it states that every action have it reaction with same magnitude but in the opposite direction.

Therefore

F = m a

a=Acceleration of the car

$a=\dfrac{F}{m}$

F= M a'

a'=Acceleration of the massive truck

$a'=\dfrac{F}{M}$

Here given that M > m that is why a > a'

Therefore car will experiences more acceleration.

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The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold 24 x (mass o

matrenka [14]

Answers:

(a) $0.0073kg$

(b) Volume gold: $3.79(10)^{-7}m^{3}$ , Volume cupper: $7.6(10)^{-8}m^{3}$

(c) $17633.554kg/m^{3}$

Explanation:

<h2>(a) Mass of gold </h2><h2 />

We are told the total mass $M$ of the coin, which is an alloy of gold and copper is:

$M=m_{gold}+m_{copper}=7.988g=0.007988kg$ (1)

Where $m_{gold}$ is the mass of gold and $m_{copper}$ is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats $K$ and mass is:

$K=24\frac{m_{gold}}{M}$ (2)

Finding ${m_{gold}$ :

$m_{gold}=\frac{22}{24}M$ (3)

$m_{gold}=\frac{22}{24}(0.007988kg)$ (4)

$m_{gold}=0.0073kg$ (5) This is the mass of gold in the coin

<h2>(b) Volume of gold and cupper</h2><h2 />

The density $\rho$ of an object is given by:

$\rho=\frac{mass}{volume}$

If we want to find the volume, this expression changes to: $volume=\frac{mass}{\rho}$

For gold, its volume $V_{gold}$ will be a relation between its mass $m_{gold}$ (found in (5)) and its density $\rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}$ :

$V_{gold}=\frac{m_{gold}}{\rho_{gold}}$ (6)

$V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}$ (7)

$V_{gold}=3.79(10)^{-7}m^{3}$ (8) Volume of gold in the coin

For copper, its volume $V_{copper}$ will be a relation between its mass $m_{copper}$ and its density $\rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}$ :

$V_{copper}=\frac{m_{copper}}{\rho_{copper}}$ (9)

The mass of copper can be found by isolating $m_{copper}$ from (1):

$M=m_{gold}+m_{copper}$

$m_{copper}=M-m_{gold}$ (10)

Knowing the mass of gold found in (5):

$m_{copper}=0.007988kg-0.0073kg=0.000688kg$ (11)

Now we can find the volume of copper:

$V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}$ (12)

$V_{copper}=7.6(10)^{-8}m^{3}$ (13) Volume of copper in the coin

<h2>(c) Density of the sovereign coin</h2><h2 />

Remembering density is a relation between mass and volume, in the case of the coin the density $\rho_{coin$ will be a relation between its total mass $M$ and its total volume $V$ :

$\rho_{coin}=\frac{M}{V}$ (14)

Knowing the total volume of the coin is:

$V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3}$ (15)

$\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}}$ (16)

Finally:

$\rho_{coin}=17633.554kg/m^{3}}$ (17) This is the total density of the British sovereign coin

6 0

3 years ago

A 540 kg satellite moves through deep space with a speed of 27 m/s. A booster rocket on the satellite fires for 1.4 s, giving a

prohojiy [21]

Answer: a

Explanation: because the answer is 1.4444444 and that's the closest

6 0

2 years ago

Read 2 more answers

Write a brief essay describing how Newton’s Laws explain how a rocket in space can move objects

statuscvo [17]

The First Law describes how an object acts when no force is acting upon it. So, rockets stay still until a force is applied to move them. Likewise, once they're in motion, they won't stop until a force is applied. Newton's Second Law tells us that the more mass an object has, the more force is needed to move it. A larger rocket will need stronger forces (eg. more fuel) to make it accelerate. The space shuttles required seven pounds of fuel for every pound of payload they carry. Newton's Third Law states that "every action has an equal and opposite reaction". In a rocket, burning fuel creates a push on the front of the rocket pushing it forward.

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3 years ago

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Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 kg and exerts

olga2289 [7]

Answer:

a) a = - 0.106 m/s^2 (←)

b) T = 12215.1064 N

Explanation:

If

F₁ = 9*1350 N = 12150 N (→)

F₂ = 9*1365 N = 12285 N (←)

∑Fx = M*a = (M₁ +M₂)*a (→)

F₁ - F₂ = (M₁ +M₂)*a

→ a = (F₁ - F₂) / (M₁ +M₂ ) = (12150-12285)N/(9*68+9*73)Kg

→ a = - 0.106 m/s^2 (←)

(b) What is the tension in the section of rope between the teams?

If we apply ∑Fx = M*a for the team 1

F₁ - T = - M₁*a ⇒ T = F₁ + M₁*a

⇒ T = 12150 N + (9 * 68 Kg)*(0.106 m/s^2)

⇒ T = 12215.1064 N

If we choose the team 2 we get

- F₂ + T = - M₂*a ⇒ T = F₂ - M₂*a

⇒ T = 12285 N - (9 * 73 Kg)*(0.106 m/s^2)

⇒ T = 12215.1064 N

4 0

3 years ago

A 46.0-kg box is being pushed a distance of 8.80 m across the floor by a force P whose magnitude is 171 N. The force P is parall

dolphi86 [110]

Answer:

a) 1504.8 J

b) 991.76 J

c) 0J

d) 0J

Explanation:

(a) The work done by the force P on the box is given by the following formula:

$W_P=Px$

P: applied force = 171N

x: distance in which the for P is applied = 8.80m

you replace the values of P and x and obtain:

$W_P=(171N)(8.80m)=1504.8J$

(b) The work don by the friction force is:

$W_f=F_fx=\mu N x=\mu Mg x$

μ = coefficient of kinetic friction = 0.250

M: mass of the box = 46.0kg

g: gravitational constant = 9.8 m/s^2

$W_f=(0.250)(46.0kg)(9.8m/s^2)(8.80m)=991.76J$

(c) The Normal force is

$N=Mg=(46.0kg)(9,8m/s^2)=450.8N$

but this force does not do work on the box because the direction is perpendicular to the direction of the force P.

$W_N=0J$

(d) the same as before:

$W_g=0J$

8 0

2 years ago