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masya89 [10]
3 years ago
13

The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of -5.7

5 m/s^2 for 4.40 s, making straight skid marks 60.0 m long, all the way to the tree. With what speed does the car then strike the tree? m/s
Physics
1 answer:
Furkat [3]3 years ago
3 0

Answer:

0.99 m/s

Explanation:

t = Time taken = 4.4 seconds

u = Initial velocity

v = Final velocity

s = Displacement = 60 m

a = Acceleration = -5.75 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{60-\frac{1}{2}\times -5.75\times 4.4^2}{4.4}\\\Rightarrow u=26.29\ m/s

v=u+at\\\Rightarrow v=26.29+(-5.75)\times 4.4\\\Rightarrow v=0.99\ m/s

The car hits the tree at 0.99 m/s

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A motorcyclist drove 7 km at 57km/h and then another 7 km at 81 km/h. What was the average speed? ​
alex41 [277]

<u>Answer:</u>

<em>The average speed of the car is 66.9 km/h</em>

<u>Explanation:</u>

Here distance covered with the speed <em>57 km/h=7 km  </em>

distance covered with the speed of <em>81 km/h=7 km</em>

<em>Average speed is equal to the ratio of total distance to the total time. </em>

<em>total distance= 7 + 7= 14 km  </em>

<em>time= \frac{distance}{speed} </em>

<em>time taken to cover the first 7 km= 7/57 h  </em>

<em>time taken to cover the second part of the journey = 7/81 h </em>

<em>average speed =  14/(7/57+7/81)=(14 \times 57 \times 81)/945=66.9 km/h</em>

<u><em>Shortcut: </em></u>

<em>When equal distances are covered with different speeds average speed=2 ab/(a+b) where a and b are the variable speeds in the phases. </em>

7 0
3 years ago
If the velocity of a long jumper during take-offat an angle of 30 degrees is 42 f/s, how fast is he moving forward (Vx) and how
Arturiano [62]

Answer:

Vx=  11.0865(m/s)

Vy=  6.4008(m/s)

Explanation:

Taking into account that 1m is equal to 0.3048 ft, the takeoff speed in m / s will be:

V= 42(ft/s) × 0.3048(m/ft) = 12.8016(m/s)

The take-off angle is equal to 30 °, taking into account the Pythagorean theorem the velocity on the X axis will be:

Vx= 12.8016 (m/s) × cos(30°)= 11.0865(m/s)

And for the same theorem the speed on the Y axis will be:

Vy= 12.8016 (m/s) × sen(30°)= 6.4008(m/s)

5 0
3 years ago
HELP BRANLIEST
Vsevolod [243]

Explanation:

- Newton's first law of motion:

"An object at rest (or in uniform motion) remains at rest (or in uniform motion) unless acted upon an unbalanced force

In this situation, we can apply Newton's first law to the keys of the keyboard that are not hit by the fingers of the man. In fact, as no force act on the keys, they remain at rest.

- Newton's second law of motion:

"The acceleration experienced by an object is proportional to the net force exerted on the object; mathematically:

F=ma

where F is the net force, m is the mass of the object, and a its acceleration"

In this case, we can apply Newton's second law to the keys of the keyboard that are hit by the man: in fact, as they are hit, they experience a downward force, and therefore they experience a downward acceleration.

"Newton's third law of motion:

"When an object A exerts a force on an object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"

Here We can apply Newton's third law to the pair of objects finger-key: in fact, as the finger apply a force on the key (action force), then the key exerts a force back on the finger (reaction force), equal and opposite.

3 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!! A IS NOT THE CORRECT ANSWER
patriot [66]

Answer: A

<u>Explanation:</u>

NOTES:

d = 650 meters

t = 10 seconds

**********************************

v = d/t

  = 650 meters/10 seconds

  = 65 meters/second

6 0
3 years ago
Read 2 more answers
A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly
Elden [556K]

Answer:

a. B= 9.45 \times10^{-3} T

b. B= 0.820 T

c. B= 0.0584 T

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

J= \frac {-I_{c}}{\pi(c^{2}-b^{2}  ) }}

J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}

B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\

B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

3 0
3 years ago
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