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3 years ago

8

A car with a mass of 1500 kg accelerates when the traffic light turns green. If the net force on the car is 3750 newtons, what i

s the car's acceleration? *
Your answer

1 answer:

timurjin [86]3 years ago

8 0

Answer:

2.5 ms⁻²

Explanation:

By Newton's 2nd law,

The rate of change of momentum is directly proportional to the unbalance force applied on the object,

By that you can get the equation,

F = ma ⇒ a = F/m

where terms are in usual meaning

a = 3750/1500 = 2.5 ms⁻²

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A man jumps from a 45 m cliff into a large body of water. How long will he fall before he enters the wate

julia-pushkina [17]

Answer:

about 2.5 seconds depending on if he jumps up and out or if he jumps out.

Explanation:

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2 years ago

A 20-kg block is held at rest on the inclined slope by a peg. A 2-kg pendulum starts at rest in a horizontal position when it is

gregori [183]

Complete Question

The diagram of this question is shown on the first uploaded image

Answer:

The distance the block slides before stopping is $d = 0.313 \ m$

Explanation:

The free body diagram for the diagram in the question is shown

From the diagram the angle is $\theta = 25 ^o$

$sin \theta = \frac{h}{d}$

Where $h = h_b - h_a$

So $d sin \theta = h_b - h_a$

From the question we are told that

The mass of the block is $m = 20 \ kg$

The mass of the pendulum is $m_p = 2 \ kg$

The velocity of the pendulum at the bottom of swing is $v_p = 15 m/s$

The coefficient of restitution is $e =0.7$

The coefficient of kinetic friction is $\mu _k = 0.5$

The velocity of the block after the impact is mathematically represented as

$v_2 f = \frac{m_b - em_p}{m_b + m_p} * v_2 i + \frac{[1 + e] m_1}{m_1 + m_2 } v_p$

Where $v_2 i$ is the velocity of the block before collision which is 0

$= \frac{20 - (0.7 * 2)}{(2 + 20)} * 0 + \frac{(1 + 0.7) * 2 }{2 + 20} * 15$

Substituting value

$v_2 f = 2.310\ m/s$

According to conservation of energy principle

The energy at point a = energy at point b

So $PE_A + KE _A = PE_B + KE_B + E_F$

Where

$PE_A$ is the potential energy at A which is mathematically represented as

$PE_A = m_b gh_a$ = 0 at the bottom

$KE _A$ is the kinetic energy at A which is mathematically represented as

$K_A = \frac{1}{2} m_b * v_2f^2$

$PE_B$ is the potential energy at B which is mathematically represented as

$PE_B = m_b gh$

From the diagram $h = h_b -h_a$

$PE_B = m_b g(h_b - h_a)$

$KE _B$ is the kinetic energy at B which is 0 (at the top )

Where is $E_F$ is the workdone against velocity which from the diagram is

$\mu_k m_b g cos 25 *d$

So

$\frac{1}{2} m_b v_2 f^2 = m_b g h_b + \mu_k m_b g cos \25 * d$

Substituting values

$\frac{1}{2} * 20 * 2.310^2 = 20 * 9.8 * d sin(25) + 0.5* 20 * 9.8 * cos 25 * d$

So

$d = 0.313 \ m$

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3 years ago

A small metal sphere weighs 0.34 N in air and has a volume of 13 cm3 . What is the acceleration of the sphere as it falls throug

Xelga [282]

To solve this problem we will apply the concepts related to the balance of forces. Said balance will be given between buoyancy force and weight, both described as derived from Newton's second law, are given as

Buoyancy force

$F_B = V\rho g$

Here,

V = Volume

$\rho$ =Density of air

g = Acceleration due to gravity

Weight

$F_W = mg$

m = mass

g = Gravity

Our values are given as,

$\text{Weight of the sphere} = W = 0.34 N$

$\text{Volume} = V = 13 cm^3 = 13*10^{-6}m^3$

$\text{density of air} =\rho =1.29kg/m^3$

$\text{gravity}= g = 9.8 m/s^2$

Then,

$F = V\rho g$

Replacing,

$F = (13*10^{-6}m^3 )(1.29Kg/m^3)( 9.8 m/s^2) = 1.6434* 10^{-4} N$

Now net force is ,

$F_{net} = mg - F$

Mass of the sphere is

$m = \frac{W}{g} = \frac{0.34N}{9.8m/s^2} = 0.03469 kg$

Now acceleration of the sphere is

$a = \frac{F_{net}}{m}$

$a = \frac{( 0.34 N)- (1.6434* 10^{-4} N)}{0.03469 kg}$

$a = 9.822m/s^2$

Therefore the acceleration of the sphere as it falls through water is $9.822m/s^2$

5 0

3 years ago