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solmaris [256]
2 years ago
14

2 Name the device in a car that we use to change the speed of the car

Physics
1 answer:
Crank2 years ago
6 0

Answer:

Speedometer

Explanation:

Have a good day! :3

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A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in m
ArbitrLikvidat [17]

Answer:

<h2>206.67N</h2>

Explanation:

The sum of force along both components x and y is expressed as;

\sum Fx = ma_x  \ and \ \sum Fy = ma_y

The magnitude of the net force which is also known as the resultant will be expressed as R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

a_x = \frac{d^2 x }{dt^2}

a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4  )\\\\a_x = \frac{d}{dt}(12t  )\\\\a_x = 12m/s^{2}

Similarly,

a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2}   )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2

\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N

R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

7 0
3 years ago
Please help me !! i’ll mark brainliest if you’re right!
navik [9.2K]
Option 2nd is the answer
3 0
2 years ago
A ball is thrown upward with an initial velocity of 13 m/s. Using the approximate value of
dem82 [27]

Answer:

v=v0 - gt

Explanation:

The equation for velocity is

v=v0 - gt

where v0=14m/s, g=10m/s^2.

in 1 second:

v=14-10=4m/s

it is positive so direction is upwards

in 2 seconds:

v=14-20=-6m/s

it is negative so direction is downwards

5 0
3 years ago
Which is a characteristic of the atom marked A
erastovalidia [21]
Its very dense. Hey, are you homeschooled?
6 0
3 years ago
Until a train is a safe distance from the station, it must travel at 5 m/s. Once the train is on open track, it can speec
nadezda [96]

Answer:

The acceleration of the train is 5 m/s².

Explanation:

Given:

let the initial velocity of a train = 5 m/s and

final velocity of a train = 45 m/s

time taken = 8 s

To find:

acceleration: ?

Solution:

We define acceleration as change in velocity per unit time that is the difference between the final velocity and initial velocity divided by time.

Acceleration = \frac{\textrm{final velocity} - \textrm{initial velocity}}{time} \\

On substituting the above values we get the required acceleration

Acceleration = \frac{45 - 5}{8}\\ =\frac{40}{8}\\ =5\ m/s^{2}

Therefore,the acceleration of the train is 5 m/s².

4 0
3 years ago
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