A crane with output power of 200W will lift a 600N object a vertical distance of 4.0 meters in seconds

Which trailer has more downward pressure where it attaches to the car?

serg [3582]

The trailer that is loaded the most. The total weight does not matter; it is about how the load is distributed. For example, our 12,000 lb snow cat trailer has weight distribution that results in less than 100 lbs of tongue weight. Heavy tongue weight can create issues, as it can shift the weight off the front wheels of the towing vehicle, causing instability.

4 0

2 months ago

A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting

serg [3582]

Response:

The speed at which the distance from the helicopter to you is changing (in ft/s) after 5 seconds is $\sqrt{725}$ ft/ sec

Clarification:

Provided:

h(t) = 25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec. 5 = 125 ft

x(5) = 10 ft/sec. 5 = 50 ft

At this point, we can determine the distance between the individual and the helicopter utilizing the Pythagorean theorem

$D(t) = \sqrt{h^2 + x^2}$

Now, let's calculate the derivative of distance in relation to time

$\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}$

By plugging in the values for h(t) and x(t) and simplifying, we arrive at,

$\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}$

$\frac{dh}{dt} = 25ft/sec$

$\frac{dx}{dt} = 10 ft/sec$

$\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}$ = $\frac{725}{\sqrt{725}}$ = $\sqrt{725}$ ft / sec

5 0

1 month ago

A block moves at 5 m/s in the positive x direction and hits an identical block, initially at rest. A small amount of gunpowder h

Yuliya22 [3333]

Answer:

Speeds of 1.83 m/s and 6.83 m/s

Explanation:

Based on the law of conservation of momentum,

$mv_o=m(v_1 + v_2)$ where m represents mass, $v_o$ is the initial speed before impact, $v_1$ and $v_2$ are the velocities of the impacted object after the collision and of the originally stationary object after the impact.

$5m=m(v_1 +v_2)$

Thus, $v_1+v_2=5$

After the collision, the kinetic energy doubles, therefore:

$2m*(0.5mv_o)=0.5m(v_1^{2}+v_2^{2})$

$2v_o^{2}=v_1^{2} + v_2^{2}$

Substituting the initial velocity of 5 m/s provides the equation needed to proceed. $v_o$

$2*(5^{2})= v_1^{2} + v_2^{2}$ We know that $v_1+v_2=5$ leads to $v_1=5-v_2$

$50=(5-v_2)^{2}+ v_2^{2}$

$50=25+v_2^{2}-10v_2+v_2^{2}$

$2v_2^{2}-10v_2-25=0$

Using the quadratic formula leads us to solve for the speeds after the explosion, specifically where a=2, b=-10, and c=-25. $v_2=6.83 m/s$

By substituting the values, the solution yields results for the speeds of the blocks, which are ultimately 1.83 m/s and 6.83 m/s.

6 0

2 months ago

10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????

serg [3582]

Answer: Tension = 47.8N, Δx = 11.5× $10^{-6}$ m.

Tension = 95.6N, Δx = 15.4× $10^{-5}$ m

Explanation: The speed of a wave on a string under tension can be determined using the following:

$|v| = \sqrt{\frac{F_{T}}{\mu} }$

$F_{T}$ denotes tension (N)

μ refers to linear density (kg/m)

Calculating the velocity:

$|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{0.00874 }$

$|v| =$ 0.0935 m/s

Distance a pulse traveled in 1.23ms:

$\Delta x = |v|.t$

$\Delta x = 9.35.10^{-2}*1.23.10^{-3}$

Δx = 11.5× $10^{-6}$

With a tension of 47.8N, the distance a pulse will cover is Δx = 11.5× $10^{-6}$ m.

When tension is doubled:

$|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{2.0.00874 }$

$|v| = \sqrt{0.01568}$

|v| = 0.1252 m/s

Distance in the same time:

$\Delta x = |v|.t$

$\Delta x = 12.52.10^{-2}*1.23.10^{-3}$

$\Delta x =$ 15.4× $10^{-5}$

With the increased tension, it moves $\Delta x =$ 15.4× $10^{-5}$ m

4 0

2 months ago

What is the internal energy (to the nearest joule) of 10 moles of Oxygen at 100 K?

serg [3582]

Response:

U = 12,205.5 J

Clarification:

To determine the internal energy of an ideal gas, use the following equation:

$U=\frac{3}{2}nRT$ (1)

U: internal energy

R: ideal gas constant = 8.135 J(mol.K)

n: number of moles = 10 mol

T: the temperature of the gas = 100K

Substituting the parameter values into equation (1):

$U=\frac{3}{2}(10mol)(8.135\frac{J}{mol.K})(100K)=12,205.5J$

The overall internal energy for 10 moles of Oxygen at 100K is 12,205.5 J

6 0

1 month ago