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4 years ago

A car travels 50 miles in 2.5 hours. Calculate the average velocity

1 answer:

5 0

In order to calculate velocity, we would need to know where the car was
when it started out, and where it was when it stopped. Without any of this
information, the best we can do is calculate its average speed.

Average speed = (distance covered) / (time to cover the distance)

   = (50 miles)    / (2.5 hours)

   =    (50 / 2.5)    mile/hour

   = 20 mile/hour .

You might be interested in

A centrifugal pump discharges 300 gpm against a head of 55 ft when the speed is 1500 rpm. The diameter of the impeller is 15.5 i

Ksivusya [100]

Answer: Question 1: Efficiency is 0.6944

Question 2: speed of similar pump is 2067rpm

Explanation:

Question 1:

Flow rate of pump 1 (Q1) = 300gpm

Flow rate of pump 2 (Q2) = 400gpm

Head of pump (H)= 55ft

Speed of pump1 (v1)= 1500rpm

Speed of pump2(v2) = ?

Diameter of impeller in pump 1= 15.5in = 0.3937m

Diameter of impeller in pump 2= 15in = 0.381

B.H.P= 6.0

Assuming cold water, S.G = 1.0

eff= (H x Q x S.G)/ 3960 x B.H.P

= (55x 300x 1)/3960x 6

= 0.6944

Question 2:

Q = A x V. (1)

A1 x v1 = A2 x V2. (2)

Since A1 = A2 = A ( since they are geometrically similar

A = Q1/V1 = Q2/V2. (3)

V1(m/s) = r x 2π x N(rpm)/60

= (0.3937x 2 x π x 1500)/2x 60

= 30.925m/s

Using equation (3)

V2 = (400 x 30.925)/300

= 41.2335m/s

To rpm:

N(rpm) = (60 x V(m/s))/2 x π x r

= (60 x 41.2335)/ 2× π × 0.1905

= 2067rpm.

6 0

3 years ago

A centrifuge in a medical laboratory rotates at an angular speed of 3,650 rev/min. when switched off, it rotates through 48.0 re

AlladinOne [14]

First of all we need to convert everything into SI units.

Let's start with the initial angular speed, $\omega _i = 3650 rev/min$ . Keeping in mind that
$1 rev = 2 \pi rad$
$1 min=60 s$
we have
$\omega _i = 3650 \frac{rev}{min} \cdot \frac{2 \pi rad/rev}{60 s/min} =382.0 rad/s$

And we should also convert the angle covered by the centrifuge:
$\theta = 48.0 rev= 48.0 rev \cdot 2 \pi \frac{rad}{rev}=301.4 rad$

This is the angle covered by the centrifuge before it stops, so its final angular speed is $\omega_f =0$ .

To solve the problem we can use the equivalent of
$2aS = v_f^2 -v_i^2$
of an uniformly accelerated motion but for a rotational motion. It will be
$2 \alpha \theta = \omega_f^2-\omega_i^2$
And by substituting the numbers, we can find the value of $\alpha$ , the angular acceleration:
$\alpha=- \frac{\omega_i^2}{2 \theta}=- \frac{(382 rad/s)^2}{2 \cdot 301.4 rad}=-242.1 rad/s^2$

4 0

3 years ago

9. A force is applied horizontally to a 20 kg box on a flat table, if the acceleration of the box is 2 m/s and the

marysya [2.9K]

By Newton's second law,

n + (-w) = 0

p + (-f ) = (20 kg) (2 m/s²)

where n is the magnitude of the normal force, w is the weight of the box, p is the magnitude of the applied force (p for push or pull), and f is the magnitude of the friction force.

Calculate the weight of the box:

w = (20 kg) (9.80 m/s²) = 196 N

Then

n = w = 196 N

and

f = µ n = 0.5 (196 N) = 98 N

Now solve for p :

p - 98 N = 40 N

p = 138 N

3 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Karolina [17]

Answer:

1.00 m/s^2

Explanation:

The problem can be solved by using Newton's second law:

$F=ma$

where

F is the net force acting on an object

m is the mass of the object

a is its acceleration

In this problem, F=35 N (the force applied by Amber) and

$m=32 kg+3 kg=35 kg$

is the mass of the wagon + the brother, so the acceleration of the wagon and the brother is

$a=\frac{F}{m}=\frac{35 N}{35 kg}=1.00 m/s^2$

5 0

4 years ago

Learning Goal: To practice Problem-Solving Strategy 7.1 Rotational dynamics problems. Suppose that you are holding a pencil bala

aliya0001 [1]

Answer:

when the pencil is balanced on a point it is motionless due to net force only acting trough the center, but when we release the position we are changing the location of a pencil hence due to turning affect it will expericence torque and the angular acceleration produced.

for the complete answer the length and mass of pencil should be known i am giving the general expression for the calculation of angular acceleration

Explanation:

<h3>τ=Iα</h3><h3>τ= torque</h3><h3>I = inertia</h3><h3>α= angular acceleration</h3>

where

I= mL^2/3 (by geometry of pencil)

when the pencil is released it will experience force due to weight

F=mgsinθ

so τ=Fxd

τ=mgsin(10°)x(L/2)

then.

α=τ/I

α=mgsin(10°)xL/2÷(mL^2)/3

6 0

3 years ago