All categories

2 years ago

Plz hurry How fast can work be done by a 7 horsepower engine?

Physics

2 answers:

den301095 [7]2 years ago

7 0

Answer:

It will go about 45-50 mph.

Explanation:

Depending on what you are hauling.

ddd [48]2 years ago

6 0

The answer is it will go at least 45-50 mph

You might be interested in

Premium

Luda [366]

Explanation:

the formula of speed is distance traveled by time it work

5 0

2 years ago

Read 2 more answers

A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta

weqwewe [10]

Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

$\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq 0 \ \ \ -------> \ \ \ 1$

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

If the value is less than zero; then we conclude that the assumption is wrong.

Then, the flow is said to be in the opposite direction

Formula for the change in specific entropy can be calculated as:

$s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \ -------> \ \ \ 2$

where;

$s_1, s_2 , s^0(T_2), s^0(T1)$ are specific entropies

R = universal gas constant

$P_1$ = pressure at location 1

$P_2$ = pressure at location 2

We obtain the specific properties of air at temperature at $T_1$ = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

$s^0(T1)$ = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature $T_2$ = 22°C + 273) K = 295 K

From the table A- 22

$s^0(T_2)$ = 1.68515 kJ/kg . K

R = $\frac{8.314 kJ}{28.97 kg.K}$

$P_1 =$ 0.95 bar

$P_2 =$ 0.8 bar

Now replacing our values into equation (2) from above; we have;

$s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )$

$s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97} \ In (\frac{0.8}{0.95} )$

$s_2-s_1 = 1.68515 -1.8279+ 0.0493$

$s_2-s_1 =-0.0934 \ kJ/kg.K$

Equating our result to equation (1)

$s_2-s_1 \geq 0\\-0.0934 \leq 0$

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

3 0

3 years ago

what is the property of an object due to its mass by which it resists any change in its position unless overcome by force ?

UkoKoshka [18]

Answer:

Mass as a Measure of the Amount of Inertia

All objects resist changes in their state of motion. All objects have this tendency - they have inertia.

Explanation:

hope this helps

5 0

2 years ago

A 0.3-kg object connected to a light spring with a force constant of 19.3 N/m oscillates on a frictionless horizontal surface. A

Ghella [55]

The total work W done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is

W = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J

That is,

• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point

• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium

so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.

By the work-energy theorem,

W = ∆K = K

where K is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So

W = 1/2 mv ²

where m is the mass of the object and v is the speed you want to find. Solving for v, you get

v = √(2W/m) ≈ 0.46 m/s

8 0

2 years ago

A woman takes her dog rover for a walk on a leash. She pulls on the leash with a force of 30.0 N at an angle of 29° above the ho

IrinaK [193]

19.8 N force is tending to lift Rover vertically off the ground.

<h3>What is horizontal and vertical component?</h3>

The horizontal velocity component ( $v_{x}$ ) describes the influence of the velocity in displacing the projectile horizontally. The vertical velocity component ( $v_{y}$ ) describes the influence of the velocity in displacing the projectile vertically.