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3 years ago

Which is a compound machine?

Physics

2 answers:

yarga [219]3 years ago

7 0

Out of the answer choices on edginuity

A) a broom

B) a screwdriver

C) a slide

D) a pair of scissors

I would say it would be the pair of scissors because of the two blades and a wedge.

remember ΩVega

Dafna1 [17]3 years ago

5 0

Compound machine<span>. A </span>machine<span> consisting of two or more simple </span>machines<span> operating together, as a wheelbarrow consisting of a lever, axle, and wheel.</span>

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How can you tell whether something has kinetic energy?

Bond [772]

<h2>If the object is moving then it has kinetic energy</h2>

6 0

3 years ago

An insulated piston–cylinder device contains 5 l of saturated liquid water at a constant pressure of 150 kpa. an electric resist

Semmy [17]

At the initial state: v1 = vf = 0.001053 m 3 /kg, h1 = hf = 467.11 kJ/kg, and s1 = sf = 1.4336 kJ/kgK.
The mass of the water is: m = V/v1 = 0.005/0.001053 = 4.7483 kg.
To find the final state, we will use the First Law:
Q12 = m(h2 - h1) for closed system undergoing a constant pressure process.
h2 = 1Q2/m + h1 = 2200/4.7483 + 467.11 = 930.43 kJ/kg.
At P2 = P1 = 150 kPa, this is a saturated mixture.
hf = 467.11 kJ/kg, hfg = 2226.5 kJ/kg, sf = 1.4336 kJ/kgK, and sfg = 5.7897 kJ/kgK
s2 = sf + sfg (h2 – hf )/hfg = 1.4336 + 5.7897(930.43 – 467.11)/2226.5 = 2.6384 kJ/kgK.
The entropy change of water is:
Delta Ssys= m(s2 – s1) = 4.7483(2.6384 – 1.4336) = 5.72 kJ/K.

8 0

3 years ago

A 2000 kg car moves along a horizontal road at speed vo

cluponka [151]

Answer:

The shortest possible stopping distance of the car is 175.319 meters.

Explanation:

In this case we see that driver use the brakes to stop the car by means of kinetic friction force. Deceleration of the car is directly proportional to kinetic friction coefficient and can be determined by Second Newton's Law:

$\Sigma F_{x} = -\mu_{k}\cdot N = m \cdot a$ (Eq. 1)

$\Sigma F_{y} = N-m\cdot g = 0$ (Eq. 2)

After quick handling, we get that deceleration experimented by the car is equal to:

$a = -\mu_{k}\cdot g$ (Eq. 3)

Where:

$a$ - Deceleration of the car, measured in meters per square second.

$\mu_{k}$ - Kinetic coefficient of friction, dimensionless.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $\mu_{k} = 0.0735$ and $g = 9.807\,\frac{m}{s^{2}}$ , then deceleration of the car is:

$a = -(0.0735)\cdot (9.807\,\frac{m}{s^{2}} )$

$a = -0.721\,\frac{m}{s^{2}}$

The stopping distance of the car ( $\Delta s$ ), measured in meters, is determined from the following kinematic expression:

$\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$ (Eq. 4)

Where:

$v_{o}$ - Initial speed of the car, measured in meters per second.

$v$ - Final speed of the car, measured in meters per second.

If we know that $v_{o} = 15.9\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $a = -0.721\,\frac{m}{s^{2}}$ , stopping distance of the car is:

$\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(15.9\,\frac{m}{s} \right)^{2}}{2\cdot \left(-0.721\,\frac{m}{s^{2}} \right)}$

$\Delta s = 175.319\,m$

The shortest possible stopping distance of the car is 175.319 meters.

8 0

4 years ago

Assuming this is a distance time graph( ignore the speed time title) assume metres on vertical scale. describe in as much detail

riadik2000 [5.3K]

Answer:

The journey started from point A, with a speed of 1 m/s for 10 seconds after which the it became stationary at 10 meters from the start point, for 10 seconds. The journey continued at a higher speed of 2 m/s for 5 seconds and halted (became stationary) again at 20 meters which was the maximum distance reached from the start point

The return journey to the start point started at the 20 meter mark and lasted for 5 seconds, at a speed of 4 m/s

The total distance travelled during the journey from the start point A to the final point B is 40 m

Explanation:

From the start point A to point B, we have;

The speed from A to B = 10 m/(10 s) = 1 m/s

The distance traveled from A to B = 10 m

The time it takes to move from A to B = 10 seconds

From the point B to point C, we have;

The distance traveled from B to C = 0 m, (stationary)

The time it remains at point B distance from the start point = 10 seconds

The speed between point B to C = 0 m/(10 s) = 0 m/s

From the point C to point D, we have;

The distance traveled from C to D = 10 m

The time it takes to move from C to D = 5 seconds

The speed between point C and D = 10 m/(5 s) = 2 m/s

From the point D to point E, we have;

The distance traveled from D to E = 0 m, (stationary)

The time it remains at point D distance from the start point = 10 seconds

The speed between point D to E = 0 m/(10 s) = 0 m/s

From the point E to point F, we have;

The distance traveled from E to F = 20 m (return journey starts at point E)

The time it takes to move from E to F = 5 seconds

The speed between point E to F = 20 m/(5 s) = 4 m/s (Return journey)

Therefore, the journey started from point A, with a speed of 1 m/s for 10 seconds after which the it became stationary at 10 meters from the start point, for 10 seconds. The journey continued at a higher speed of 2 m/s for 5 seconds and halted (became stationary) again at 20 meters which was the maximum distance reached from the start point

The return journey to the start point started at the 20 meter mark and lasted for 5 seconds, at a speed of 4 m/s

The total distance moved, 'd', to and from the start point with reference to the graph is given as follows;

d = (From A to B) 10 m + (From B to C) 0 m + (From C to D) 10 m + (From D to E) 0 m + (From E to F) 20 m = 40 m

The total distance travelled in the journey is 40 m

The total displacement, $\underset{d}{\rightarrow}$ = 10 m + 10 m - 20 m = 0 m

7 0

3 years ago

What do we call the substances that react together (break/form chemical bonds) to form new substances?

Luda [366]

Answer: A) reactants

Explanation:

Reactants are the substances that have chemical reactions when they are combined, whereas the product is what comes from the reactants’ chemical reactions.

3 0

4 years ago

Read 2 more answers