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2 years ago

12

A hockey puck is pushed by a stick with a force of 750 newtons. The puck travels 2.0 meters in 0.30 seconds. How powerful is the

push? (Ignore frictional effects.)

1 answer:

nekit [7.7K]2 years ago

5 0

The work done by the force pushing the puck is equal to the product between the force and the distance covered:
$W=Fd=(750 N)(2.0 m)=1500 J$

And the power is the work done per unit time:
$P= \frac{W}{t}$
and using t=0.30 s, we find the power:
$P= \frac{1500 J}{0.30 s}=5000 W$

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A 0.3 mm long invertebrate larva moves through 20oC water at 1.0 mm/s. You are creating an enlarged physical model of this larva

AleksandrR [38]

Answer:

Explanation:

For the problem, we should have same reynolds number

ρvd/mu = constant

1000×1×10⁻³×0.3×10⁻³/1.002×10⁻³ = 1400×0.5×d/600

d = 25.66 cm

5 0

2 years ago

 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer

Andrew [12]

Answer:

a) the mug hits the floor 0.7425m away from the end of the bar. b) |V|=5.08m/s θ= -72.82°

Explanation:

In order to solve this problem, we must first start by doing a drawing of the situation. (see attached picture).

a)

From the drawing we can see that we are dealing with a two dimensions movement problem. So in order to find out how far away from the bar the mug will fall, we need to start by finding how long it will take the mug to be in the air, so we analyze the vertical movement of the mug.

In order to find the time we need to use the following formula, which contains the data we know:

$y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}$

we know that $y_{f}=0$ and that $v_{y0}=0$ as well, so the formula is simplified to:

$0=y_{0}+\frac{1}{2}at^{2}$

we can now solve this for t, so we get:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

we know that $y_{0}=1.20m$ and that $a=g=-9.8m/s^{2}$

the acceleration of gravity is negative because the mug is moving downwards. So we substitute them into the given formula:

$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

which yields:

t=0.495s

we can now use this to find the horizontal distance the mug travels. We know that:

$V_{x}=\frac{x}{t}$

so we can solve this for x, so we get:

$x=V_{x}t$

and we can now substitute the values we know:

x=(1.5m/s)(0.495s)

which yields:

x=0.7425m

b) Now that we know the time it takes the mug to hit the floor, we can use it to find the final velocity in the y-direction by using the following formula:

$a=\frac{v_{f}-v_{0}}{t}$

we know the initial velocity in the vertical direction is zero, so we can simplify the formula:

$a=\frac{v_{f}}{t}$

so we can solve this for the final velocity:

$V_{yf}=at$

in this case the acceleration is the same as the acceleration of gravity (which is negative) so we can substitute that and the time we found on the previous part to get:

$V_{yf}=(-9.8m/s^{2})(0.495s)$

which yields:

$V_{yf}=-4.851m/s$

so now we know the components of the final velocity, which are:

$V_{xf}=1.5m/s$ and $V_{yf]=-4.851m/s$

so now we can find the speed by determining the magnitude of the vector, like this:

$|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}$

so we get:

$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

which yields:

|V|=5.08m/s

now, to find the direction of the impact, we can use the following equation:

$\theta = tan^{-1} (\frac{V_{y}}{V_{x}})$

so we get:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

which yields:

$\theta = -72.82^{o}$

4 0

2 years ago

If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is $u = 7.73 \ m/s$

Explanation:

From the question we are told that

The height at which he let go of the brief case is h = 130 m

The time taken before the the brief case hits the water is t = 6 s

Generally the initial speed of the briefcase (Which also the speed of the helicopter )before the man let go of it is mathematically evaluated using kinematic equation as

$s = h+ u t + 0.5 gt^2$

Here s is the distance covered by the bag at sea level which is zero

$0 = 130+ u * (6) + 0.5 * (-9.8) * (6)^2$

=> $0 = 130+ u * (6) + 0.5 * (-9.8) * (6)^2$

=> $u = \frac{-130 + (0.5 * 9.8 * 6^2) }{6}$

=> $u = 7.73 \ m/s$

7 0

2 years ago

A horizontal spring with spring constant 85 n/m extends outward from a wall just above floor level. a 3.5 kg box sliding across

Rina8888 [55]

k = spring constant of the spring = 85 N/m

m = mass of the box sliding towards the spring = 3.5 kg

v = speed of box just before colliding with the spring = ?

x = compression the spring = 6.5 cm = 6.5 cm (1 m /100 cm) = 0.065 m

the kinetic energy of box just before colliding with the spring converts into the spring energy of the spring when it is fully compressed.

Using conservation of energy

Kinetic energy of spring before collision = spring energy of spring after compression

(0.5) m v² = (0.5) k x²

m v² = k x²

inserting the values

(3.5 kg) v² = (85 N/m) (0.065 m)²

v = 0.32 m/s

8 0

2 years ago

Argon in the amount of 1.5 kg fills a 0.04-m3 piston cylinder device at 550 kPa. The piston is now moved by changing the weights

Arlecino [84]

Answer:

275 kPa

Explanation:

mass of the gas=m=1.5 kg

initial volume if the gas=V₁=0.04 m³

initial pressure of the gas= P₁=550 kPa

as the condition is given final volume is double the initial volume

V₂=final volume

V₂=2 V₁

As the temperature is constant

T₁=T₂=T

$\frac{P1V1}{T1}$ = $\frac{P2 V2}{T2}$

putting the values in the equation.

$\frac{P1V1}{T1}$ = $\frac{P2 *2V1}{T2}$

P₂= $\frac{P1}{2}$

P₂= $\frac{550}{2}$

P₂=275 kPa

So the final pressure of the gas is 275 kPa.

3 0

2 years ago