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3 years ago

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um carro com uma velocidade de 80 km\h passa pelo km 240 de uma rodovia as 7:30. a que horas este carro chegara a proxima cidade

, sabendo-se que a mesma esta situada no km 300 dessa rodovia?

1 answer:

Shkiper50 [21]3 years ago

3 0

From km 240 to km 300 there are 60 km.

The time to travel 60 km at 80km/h is t = d/V = 60km / 80 km/h = 3/4 h.

That is 45 minutes. Then the car will arrive at 7:30 + 45 min = 8:15.

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A toy spacecraft is launched directly upward. When the toy reaches its highest point, a spring is released and the toy splits in

Murljashka [212]

Answer:

A

Explanation:

Momentum conservation will cause 0.08kg to move to the west (opposite of 0.02 kg).

and because both are at the same height above the ground, they will take the same time to reach the ground.

The speed of 0.08kg will be less than 0.02 kg, let v be the speed of 0..02kg, then speed of 0.08kg V is

0.02v - (0.08)V = 0

V = 0.02 v/ 0.08 = v/4

The speed of 0.08 kg = v/4

The speed of 0.08 kg is less than 0.02kg.

So 0.02kg strikes the ground farther from the launch point than does the 0.08 kg

8 0

3 years ago

A 0.86 kg rock is projected from the edge of the top of a building with an initial velocity of 8.65 m/s at an angle 46◦ above th

Georgia [21]

Answer:

Height of the building = 11.4 m

Explanation:

As we know that the stone is projected at an angle 46 degree with speed 8.65 m/s

so the two components of the speed is given as

$v_x = 8.65 cos46$

$v_x = 6 m/s$

vertical component of the speed is given as

$v_y = 8.65 sin46$

$v_y = 6.22 m/s$

now we know that the ball strike at horizontal distance of 13.7 m

so we will have

$x = v_x t$

$13.7 = 6 t$

$t = 2.28 s$

now we know that in vertical direction ball will move under uniform gravity so we can use kinematics

$y = v_y t + \frac{1}{2}at^2$

$y = 6.22(2.28) - \frac{1}{2}(9.81)(2.28^2)$

$y = -11.4 m$

Height of the building = 11.4 m

3 0

3 years ago

Read 2 more answers

Describe all the ways that newtons laws can apply in a car crash

Harman [31]

Newtons first law - Objects in the car at rest (The human) will remain at rest unless affected by an unbalanced force. Well the unbalanced force would be the crash and this would set the human in motion and they would ether fly out the car if not wearing a seat belt or if wearing one they would get bad whip lash

Newtons second law - With more mass requires more force, so since the human is pretty light or even if heavy in a big crash there will be so much more from it that this will send the human flying.

Newtons 3rd law - Objects A puts force onto objects b and object b excretes the same amount of force back onto object a, so in a crash the human would hit the car hard and the car would excrete the same amount of force back on the human which would really damage him/her

7 0

3 years ago

a 282 kg bumper car moving right at 3.50 m/s collides with a 155 kg bumper car moving 1.88 m/s left. afterwards, the 282 kg car

Vaselesa [24]

The momentum of the 155 kg car afterwards is 469.7 kg m/s to the right

Explanation:

We can solve the problem by using the law of conservation of momentum: the total momentum of the system is conserved before and after the collision, so we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 282 kg$ is the mass of the bumper car

$u_1 = 3.50 m/s$ is the initial velocity of the bumper car (we take the right as positive direction)

$v_1 = 0.800 m/s$ is the final velocity of the bumper car

$m_2 = 155 kg$ is the mass of the second bumper car

$u_2 = -1.88 m/s$ is the initial velocity of the second car (moving to the left)

$v_2$ is the final velocity of the second car

Solving for $v_2$ ,

$v_2 = \frac{m_1 u_1+m_2 u_2 - m_1 v_1}{m_2}=\frac{(282)(3.50)+(155)(-1.88)-(282)(0.800)}{155}=3.03 m/s$

where the positive sign means the direction is to the right.

And now we can find the momentum of the 155 kg afterwards, which is

$p_2 = m_2 v_2 = (155)(3.03)=469.7 kg m/s$ (to the right)

Learn more about momentum:

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#LearnwithBrainly

7 0

3 years ago

Calculate the phase angle (in radians) for a circuit with a maximum voltage of 12 V and w-50 Hz. The voltage source is connected

Vinvika [58]

Answer:

The phase angle is 0.0180 rad.

(c) is correct option.

Explanation:

Given that,

Voltage = 12 V

Angular velocity = 50 Hz

Capacitance $C= 20\times10^{-2}\ F$

Inductance $L=20\times10^{-3}\ H$

Resistance $R= 50\ Omega$

We need to calculate the impedance

Using formula of impedance

$z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}$

$z=\sqrt{50^2+(50\times20\times10^{-3}-\dfrac{1}{50\times20\times10^{-2}})^2}$

$z=50.00$

We need to calculate the phase angle

Using formula of phase angle

$\theta=\cos^{-1}(\dfrac{R}{z})$

$\theta=\cos^{-1}(\dfrac{50}{50.00})$

$\theta=0.0180\ rad$

Hence, The phase angle is 0.0180 rad.

3 0

3 years ago