How do you calculate the rate of change of velocity

What type of acceleration does an object moving with constant speed in a circular path experience?Select one:A) constant acceler

olga_2 [115]

ANSWER:

D) centripetal acceleration.

STEP-BY-STEP EXPLANATION:

When a body performs a uniform circular motion, the direction of the velocity vector changes at every instant. This variation is experienced by the linear vector, due to a force called centripetal, directed towards the center of the circumference that gives rise to the centripetal acceleration.

Therefore, the answer is centripetal acceleration.

3 0

1 year ago

A plane flies from basecamp the Lake A, 280 km away in a direction of 20° north of

klasskru [66]

Answer:

308657

Explanation:

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5 0

3 years ago

A 1.5-kg ball is throw at 10 m/s. What is the balls momentum?

alexdok [17]

Answer : The momentum of ball is, 15 kg.m/s

Explanation :

Momentum : It is defined as the motion of a moving body. Or it is defined as the product of mass of velocity of an object.

Formula of momentum is:

where,

p = momentum = ?

m = mass = 1.5 kg

v = velocity = 10 m/s

Now put all the given values in the above formula, we get:

Therefore, the momentum of ball is 15 kg.m/s

3 0

3 years ago

A long iron bar lies along the x-axis and has current of I = 16.4 A running through it in the +x-direction. The bar is in the pr

Gre4nikov [31]

Answer:

B = 8.0487mT

Explanation:

To solve the exercise it is necessary to take into account the considerations of the Magnetic Force described by Faraday,

The magnetic force is given by the formula

$F =BILsin\theta$

Where,

B = Magnetic Field

I = Current

L = Length

$\theta =$ Angle between the magnetic field and the velocity, for this case are perpendicular, then is 90 degrees

According to our data we have that

I = 16.4A

F = 0.132N/m

As we know our equation must be modificated to Force per length unit, that is

$\frac{F}{L} = BI sin(90)$

Replacing the values we have that

$0.132 = 16.4 (1) B$

Solving for B,

$B = \frac{0.132}{16.4}$

$B = 8.0487mT$

8 0

4 years ago

a man drags a 8.10 kg bag of mulch at a constant speed, applying a 29.5 N at 38°. what is the coefficient of friction?

lesantik [10]

Answer:

The coefficient of friction is 0.38.

Explanation:

The free body diagram is drawn below.

Let $f$ be frictional force acting in the backward direction as shown. Let the coefficient of friction be $\mu$ . Let $N$ be the normal reaction force acting on the bag.

Given:

Mass of the bag is, $m=8.10\textrm{ kg}$

Force acting at $\theta = 38$ ° is $F= 29.5\textrm{ N}$

Acceleration due to gravity is, $g=9.8\textrm{ }m/s^{2}$

The force F can be resolved into its components as $F_{x}=F \cos \theta$ and $F_{y}=F \sin \theta$

Therefore,

$F_{x}=29.5\cos(38)=23.25\textrm{ N}\\F_{y}=29.5\sin(38)=18.16\textrm{ N}$

Now, as there is no acceleration in vertical direction, therefore,

Sum of upward forces = Sum of downward forces

$N+F_{y}=mg\\N=mg-F_{y}=8.10\times 9.8-18.16\\N=79.38-18.16=61.22\textrm{ N}$

Now, as the bag is moving at a constant speed, so acceleration in the horizontal direction is also zero as acceleration is the rate of change of velocity.

Therefore, backward force = forward force.

$f=F_{x}\\f=23.25\textrm{ N}$

Now, frictional force is given as:

$f=\mu N\\\mu = \frac{f}{N}=\frac{23.25}{61.22}=0.38$

Therefore, the coefficient of friction is 0.38.

8 0

3 years ago