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2 years ago

If the volume of an object is reported as 5.0 ft3 what is the volume in cubic meters

1 answer:

3 0

The problem statement is simply asking us to convert units. We convert from units of ft^3 to units of m^3. To do this, we need a conversion factor. For this case, we use 1 m is equal to 3.28084 ft. We do as follows:

5.0 ft^3 ( 1 m / 3.28084 ft )^3 = 0.1416 m^3

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A ball is dropped from the top of a building.After 2 seconds, it’s velocity is measured to be 19.6 m/s. Calculate the accelerati

zlopas [31]

Answer:

acceleration, a = 9.8 m/s²

Explanation:

'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.

u = 0 m/s

After 2 seconds, velocity of the ball is 19.6 m/s.

t = 2s, v = 19.6 m/s

Using

v = u + at

19.6 = 0 + 2a

a = 9.8 m/s²

6 0

2 years ago

Read 2 more answers

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

DENIUS [597]

Answer:

$v_{oy}=11.29\ m/s$

$y=7.39\ m$

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle $\theta$ with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

Where $y_o$ is the initial height above the ground level, $v_{oy}$ is the vertical component of the initial velocity and t is the time

The y-component of the speed is

$v_y=v_{oy}-gt$

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of $v_o$

The object will reach the maximum height when $v_y=0$ . It allows us to compute the time to reach that point

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum heigh is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Solving for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Replacing the known values

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

5 0

2 years ago

Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci

Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

$\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W$

$\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})$

$h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}$

$2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]$

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99

6 0

2 years ago

Rank the following situations according to the magnitude of the impulse of the net force, from largest value to smallest value.

wolverine [178]

Answer:

I and II

III and IV

Explanation:

The impulse is equal to the change in momentum of the object involved, so we can calculate the change in momentum in each situation and compare them all.

Taking always east as positive direction, and labelling

u the initial velocity

v the final velocity

m = 1000 kg the mass (which is always equal)

We find:

(i)

u = 25 m/s

v = 0

$|I|=m(v-u)=(1000)(0-25)=25,000 Ns$

(II)

u = 25 m/s

v = 0

$|I|=m(v-u)=(1000)(0-25)=25,000 Ns$

(III)

In this case,

F = 2000 N is the force

$\Delta t = 10 s$ is the time

So the magnitude of the impulse is

$|I| =F\Delta t = (2000N)(10)=20,000 Ns$

(IV)

F = 2000 N is the force

$\Delta t = 10 s$ is the time

So the magnitude of the impulse is

$|I| =F\Delta t = (2000N)(10)=20,000 Ns$

(V)

u = 25 m/s

v = -25 m/s

$|I|=m(v-u)=(1000)(-25-25)=50,000 Ns$

So the ranking from largest to smallest is:

I and II

III and IV

5 0

2 years ago

A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, w

Aloiza [94]

Answer:

25.82 m/s

Explanation:

We are given;

Force exerted by baseball player; F = 100 N

Distance covered by ball; d = 0.5 m

Mass of ball; m = 0.15 kg

Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.

We should note that work done is a measure of the energy exerted by the baseball player.

Thus;

F × d = ½mv²

100 × 0.5 = ½ × 0.15 × v²

v² = (2 × 100 × 0.5)/0.15

v² = 666.67

v = √666.67

v = 25.82 m/s

4 0

1 year ago