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goblinko [34]
3 years ago
10

You are driving your car, and the traffic light ahead turns red. you apply the brakes for 2.26 s, and the velocity of the car de

creases to + 5.43 m/s. the carâs deceleration has a magnitude of 2.78 m/s2 during this time. what is the carâs displacement? x =
Physics
1 answer:
Lorico [155]3 years ago
6 0

Let the initial velocity of the car be u.

Final velocity of the car (v) = 5.43 m/s

deceleration (a) = - 2.78 m/s^2

Time taken (t) = 2.26 s

Using the first equation of motion:

v = u + at

u = v - at

u = 5.43 - (-2.78 \times 2.26)

u = 5.43 + 6.28

u = 11.71 m/s

Let the car's displacement be x.

Using second equation of motion:

x = ut + \frac{1}{2}at^2

x = 11.71 \times 2.26 - \frac{1}{2} \times 2.78 \times 2.26^2

x = 26.4646 - 7.0995

x = 19.3651 meters

Hence, the displacement of the car is 19.36 meters

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A 0.03-kg bullet is fired with a horizontal velocity of 470 m/s and becomes embedded in block B which has a mass of 3 kg. After
Gala2k [10]

Answer with Explanation:

We are given that

Mass of bullet,m_1=0.03 kg

u_1=470 m/s

m_2=3 kg

\mu_k=0.2

m_3=30 kg

We have to find the velocity of the bullet and block B after the first impact and final velocity of the carrier.

According to law of conservation of momentum

m_1u_1=(m_1+m_2)v

0.03(470)=(0.03+3)v

v=\frac{0.03(470)}{(0.03+3)}=4.65 m/s

Hence, the velocity of the bullet and block B after the first impact=4.65 m/s

According to law of conservation of momentum

(m_1+m_2)v=(m_1+m_2+m_3)V

(0.03+3)\times 4.65=(0.03+3+30)V

V=\frac{(0.03+3)\times 4.65}{(0.03+3+30)}

V=0.43 m/s

3 0
4 years ago
Read 2 more answers
Motor oil, with a viscosity of 0.25 N ∙ s/m2, is flowing through a tube that has a radius of 3.0 mm and is 1.0 m long. The drop
Alex73 [517]

Answer:

The average speed of the oil is 1 m/s.

Explanation:

Given that,

Viscosity \eta= 0.25 N-s/m^2

Radius r = 3.0 mm

Length = 1.0 m

Pressure = 200 kPa

We need to calculate the average speed of the oil

Using formula of pressure

\Delta P=\dfrac{8\pi\eta\rho v}{A}

v=\dfrac{A\times\Delta P}{8\pi\eta\rho}

Where, A = area

\rho = density of oil

\eta = viscosity

Put the value into the formula

v=\dfrac{\pi\times(3.0\times10^{-3})^2\times200\times10^{3}}{8\pi\times0.25\times0.9}

v=1\ m/s

Hence, The average speed of the oil is 1 m/s.

6 0
4 years ago
You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh
g100num [7]

To solve this problem we will apply trigonometric and optical concepts that allow us to obtain the minimum distance required. The resolution of the eye is given under the following condition,

\theta = \frac{1.22\lambda}{D}

Here,

\lambda = Wavelength

D = Diameter

With the values we have that the diameter will be,

\theta = \frac{1.22(534nm)}{5.37mm}

\theta = 1.213*10^{-4}

The relation between the distance of the lights and the distance from the eye to the lamp is given under the function,

sin\theta = \frac{d}{L}

For small angles sin\theta = \theta, then

\theta = \frac{d}{L}

Here,

d = Distance between lights

L = Distance from eye to lamp

1.213*10^{-4} = \frac{0.673m}{L}

L = \frac{0.673m}{1.213*10^{-4}}

L = 5548.22m

Therefore the distance will be 5.5km

5 0
3 years ago
Short, difficult activities that push your body are called
Reika [66]

Answer: A

Explanation: Any short-duration exercise that is powered primarily by metabolic pathways that do not use oxygen. Examples

of anaerobic exercise include sprinting and weight lifting.

4 0
2 years ago
Read 2 more answers
A 62.6-kg skateboarder starts out with a speed of 2.18 m/s. He does 113 J of work on himself by pushing with his feet against th
n200080 [17]

Answer:

a. Wgra=786.09J

b. 1.28m

Explanation:

The change in the potential energy is the work done by the gravitational force.

For this problem you have to take into account that the total work done is given by the change in the kinetic energy

W_{tot}=\Delta E_k=\frac{m}{2}(v_f^2-v_0^2)\\W_{tot}=\frac{62.6kg}{2}((6.55\frac{m}{s})^2-(2.18\frac{m}{s})^2)=1194.09J

Furthermore the total work is the contribution of the work done by the skater, the gravitational force and the friction

W_{tot}=W_{ska}+W_{fric}+W_{gra}

(a) by separating Wfric you have

W_{gra}=W_{tot}-W_{fric}-W_{ska}=1194.09J-295J-113J=786.09J

(b) It is only necessary to use the expression for the work done by gravitational force

W_{grav}=mgh\\h=\frac{W_{grav}}{mg}=\frac{786.09J}{(62.6kg)(9.8\frac{m}{s^2})}=1.28m

HOPE THIS HELPS!!

6 0
3 years ago
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