You push the ball with aforce of 22.8N which induces a -2.3N frictional force. What is the net force while you push?

A student is trying to calculate the density of a can. He already knows the mass, but he needs to determine the volume as well.

elena-14-01-66 [18.8K]

Answer:

V=Bh

Explanation:

B h is used for rectangular solids and cylinders

6 0

3 years ago

Which of the following pieces of laboratory equipment is not directly used to make measurements

RoseWind [281]

I believe the answer is a test tube.

4 0

3 years ago

Read 2 more answers

An object starts from rest at time t = 0.00 s and moves in the +x direction with constant acceleration. The object travels 14.0

Reptile [31]

Answer:

28 m/s^2

Explanation:

distance, s = 14 m

time, t = 2 - 1 = 1 s

initial velocity, u = 0 m/s

Let a be the acceleration.

Use third equation of motion

$s = ut + \frac{1}{2}at^{2}$

$14 = 0 + \frac{1}{2}a\times 1^{2}$

a = 28 m/s^2

Thus, the acceleration is 28 m/s^2.

7 0

3 years ago

At noon, ship A is 140 km west of ship B. Ship A is sailing east at 30 km/h and ship B is sailing north at 25 km/h. How fast (in

Luba_88 [7]

Answer:

The rate of change of distance between the two ships is 18.63 km/h

Explanation:

Given;

distance between the two ships, d = 140 km

speed of ship A = 30 km/h

speed of ship B = 25 km/h

between noon (12 pm) to 4 pm = 4 hours

The displacement of ship A at 4pm = 140 km - (30 km/h x 4h) =

140 km - 120 km = 20 km

(the subtraction is because A is moving away from the initial position and the distance between the two ships is decreasing)

The displacement of ship B at 4pm = 25 km/h x 4h = 100 km

Using Pythagoras theorem, the resultant displacement of the two ships at 4pm is calculated as;

r² = a² + b²

r² = 20² + 100²

r = √10,400

r = 101.98 km

The rate of change of this distance is calculated as;

r² = a² + b²

r = 101.98 km, a = 20 km, b = 100 km

$2r(\frac{dr}{dt} ) = 2a(\frac{da}{dt} ) + 2b(\frac{db}{dt} )\\\\r(\frac{dr}{dt} ) = a(\frac{da}{dt} ) + b(\frac{db}{dt} )\\\\101.98(\frac{dr}{dt} ) = 20(-30 ) + 100(25 )\\\\101.98(\frac{dr}{dt} ) = -600 + 2,500\\\\101.98(\frac{dr}{dt} ) = 1900\\\\\frac{dr}{dt} = \frac{1900}{101.98} = 18.63 \ km/h$

5 0

3 years ago

La trayectoria de vuelo del helicóptero cuando despega desde A esta definida por las ecuaciones x=2t^2 m y y=0.04t^3 m, donde t

Yuri [45]

Answer:

The flight path of the helicopter when it takes off from A is defined by the equations x = 2t ^ 2 m and y = 0.04t ^ 3 m, where t is the time in seconds. Determine the distance the helicopter is from point A and the magnitudes of its speed and acceleration when t = 10 s

Explanation:

Given that,

The path curve are

x = 2t²

y = 0.04t³

Distance from A. When t = 10s

The position on the x axis at t=10s is

x = 2t²

x = 2×10² = 200m

The position on y axis at t= 10s is

y = 0.04t³

y = 0.04 ×10³ = 40m

So, the plane it at (200,40) and it started for (0,0)

From geometry we can find distance between the two point using

D = √(x2-x1)² + (y2-y1)²

D = √(200-0)²+(40-0)²

D = √200²+40²

D = 203.96m

So, the distance is approximately 204m

b. The magnitude of the speed.

Speed is give as the differentiation of distance

So, for x curve, at t =10

Vx = dx/dt = 4t

Vx = 4×10 = 40m/s

Also, for y axis at t=10

Vy = dy/dt = 0.12t²

Vy = 0.12×10² = 12m/s

Then, the velocity at 10s is

V = √(Vx²+Vy²)

V = √40²+12²

V = 41.76m/s

The velocity helicopter at t=10s is 41.76m/s

c. The acceleration at t=10s?

We need to find acceleration of the curve and it is given as

a = d²x/dt²

So, for x axis at t=10

ax= d²x/dt² = 4

ax = 4m/s²

Also, for y axis at t=10

ay= d²y/dt² = 0.24t

ay = 0.24 ×10 = 2.4m/s²

Then, the magnitude of acceleration at t=10s is

a = √(ay²+ax²)

a = √2.4²+ 4²

ay = 4.66m/s²

The acceleration of the helicopter at t=10s is 4.66m/s²

4 0

3 years ago