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4 years ago

5

A 1100-kg car pulls a boat on a trailer. (Enter the magnitude.) (a) What total force (in N) resists the motion of the car, boat,

and trailer, if the car exerts a 1860-N force on the road and produces an acceleration of 0.600 m/s2

1 answer:

Finger [1]4 years ago

5 0

Answer:

780 N

Explanation:

The combined mass of boat and trailer is, m= 700 kg.

Force applied by car, F= 1860 N

acceleration produced on the total system, a = 0.600 m/s²

Apply Newton's second law:

$F-F_r=m_na$

$m_n=700+1100 =1800 kg$

Substitute the values:

$1860-F_r=1800\times0.6\\F_r=780 N$

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A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of

Veseljchak [2.6K]

Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

$h=ut-\dfrac{1}{2}gt^2$

$h=(u\sin60)\times t-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2$

$h=41.6\ m$

(b). We need to calculate the maximum height of the ball

Using formula of height

$h_{max}=\dfrac{(u\sin\theta)^2}{2g}$

Put the value into the formula

$h=\dfrac{(33\sin60)^2}{2\times 9.8}$

$h=41.67\ m$

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

$v=u-gt$

$v=u\sin\theta-gt$

Put the value into the formula

$v_{y}=33\times\sin 60-9.8\times3.0$

$v_{y}=-0.82\ m/s$

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=33\times\cos60$

$v_{x}=16.5\ m/s$

We need to calculate the ball's impact speed

Using formula of velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(16.5)^2+(-0.82)^2}$

$v=16.52\ m/s$

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

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3 years ago

What is one way that fission and fusion are similar?

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3 years ago

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A large helium filled balloon is used as the center piece for a graduation party. The balloon alone has a mass of 225 kg and it

Orlov [11]

To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and sum of forces in bodies.

From Newton's second law we understand that

$F= ma (\rightarrow$ Gravity at this case)

Where,

m = mass

a= acceleration

Also we know that

$\rho = \frac{m}{V} \Rightarrow m = \rho V$

Part A) The buoyant force acting on the balloon is given as

$F_b = ma$

As mass is equal to the density and Volume and acceleration equal to Gravity constant

$F_b = \rho V g$

$F_b = 1.2*323*9.8$

$F_b = 3798.5$

PART B) The forces acting on the balloon would be given by the upper thrust force given by the fluid and its weight, then

$F_{net} = F_b -W$

$F_{net} = F_b -(mg+\rho_H Vg)$

$F_{net} = 3798.5-(9.8*225*9.8*0.179*323)$

$F_{net} = 1030N$

PART C) The additional mass that can the balloon support in equilibrium is given as

$F_{net} = m' g$

$m' =\frac{F_{net}}{g}$

$m' = \frac{1030}{9.8}$

$m' = 105Kg$

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4 years ago

A solenoid that is 78.8 cm long has a cross-sectional area of 15.9 cm2. There are 914 turns of wire carrying a current of 8.25 A

Harlamova29_29 [7]

Answer:

(a) Energy density will be equal to $57.31J/m^3$

(b) Total energy will be equal to 0.0718 J

Explanation:

It is given that length of solenoid l = 78.8 cm = 0.788 m

Cross sectional area $A=15.9cm^2=15.9\times 10^{-4}m^2$

Number of turns of the wire N = 914

Current in the solenoid i = 8.25 A

Inductance of the wire is equal to $L=\frac{\mu _0N^2A}{l}=\frac{4\pi \times 10^{-7}\times 914^2\times 15.9\times 10^{-4}}{0.788}=2.117\times 10^{-3}H$

(b) Total energy stored in magnetic field $U=\frac{1}{2}Li^2=\frac{1}{2}\times 2.11\times 10^{-3}\times 8.25^2=0.0718J$

(a) Energy density will be equal to

$U_b=\frac{0.0718}{15.9\times 10^{-4}\times 0.788}=57.31J/m^3$

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3 years ago

car 1 is traveling south at 18 m/s and has a full load, giving it a total mass of 14,650 kg. Car 2 is traveling north at 11 m/s

kvasek [131]

Answer:

v_{4}= 80.92[m/s] (Heading south)

Explanation:

In order to calculate this problem, we must use the linear moment conservation principle, which tells us that the linear moment is conserved before and after the collision. In this way, we can propose an equation for the solution of the unknown.

ΣPbefore = ΣPafter

where:

P = linear momentum [kg*m/s]

Let's take the southward movement as negative and the northward movement as positive.

$-(m_{1}*v_{1})+(m_{2}*v_{2})=-(m_{1}*v_{3})+(m_{2}*v_{4})$

where:

m₁ = mass of car 1 = 14650 [kg]

v₁ = velocity of car 1 = 18 [m/s]

m₂ = mass of car 2 = 3825 [kg]

v₂ = velocity of car 2 = 11 [m/s]

v₃ = velocity of car 1 after the collison = 6 [m/s]

v₄ = velocity of car 2 after the collision [m/s]

$-(14650*18)+(3825*11)=(14650*6)-(3825*v_{4})\\v_{4}=80.92[m/s]$

4 0

3 years ago