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2 years ago

Before the student releases the cart by cutting the tinsel string, what forces are acting on the cart?

Physics

1 answer:

Naya [18.7K]2 years ago

5 0

Answer: TENSION and WEIGHT

Explanation:

Force experienced by the spring is called TENSION while the WEIGHT is the gravitational pull on the body towards the earth surface. Therefore the forces acting on the cart are TENSION and WEIGHT(weight acts downwards (along negative y-axis) while the TENSION upward(along positive y-axis).

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A wave has a frequency of 46 Hz and a wavelength of 1.7 meters. What is the wave speed wave?

arlik [135]

Answer:

Explanation:

(1.7 m/cycle)(46 cycle/s) = 78.2 m/s

4 0

2 years ago

The helicopter in the drawing is moving horizontally to the right at a constant velocity. The weight of the helicopter is W=4900

Sedaia [141]

Answer:

(a) The magnitude of the lift force is 52144.71 N, approximately.

(b) The magnitude of the air resistance force opposing the movement is 17834.54 N, approximately.

Explanation:

Since the helicopter is moving horizontally at a constant velocity, we can assume that the net force acting on it is zero, then

(a) in the vertical direction we have

$L\cos(20\deg)-W=0\\L=\frac{W}{\cos(20\deg)}=\frac{49000 N}{\cos(20\deg)}\approx \mathbf{52144.71 N}$ .

(b) Now horizontally,

$L\sin(20\deg)-R=0\\R=L\sin(20\deg)=52144.71 N\times \sin(20\deg) \approx \mathbf{17834.54 N}.$

3 0

2 years ago

Two speedboats are traveling at the same speed relative to the water in opposite directions in a moving river. An observer on th

DENIUS [597]

Answer:

a) vboat = 5.95 m/s b) vriver= 1.05 m/s

Explanation:

a) As observed from the shore, the speed of the boats can be expressed as the vector sum, of the boat speed relative to the water and the river speed relative to the shore, as follows:

vb₁s = vb₁w + vrs

In one case, the boat is moving in the same direction as the water:

vb₁s = vb₁w + vrs = 7.0 m/s (1)

For the other boat, it is clear that is moving in an opposite direction:

vb₂s = vb₂w - vrs = 4.9 m/s (2)

As we know that vb₁w = vb₂w, adding both sides, we can remove the river speed from the equation, as follows:

vb₁w = vb₂w = $\frac{7.0 m/s + 4.9 m/s}{2} =5.95 m/s$

b) Replacing this value in (1) and solving for vriver, we have:

vriver = 7.0 m/s - 5.95 m/s = 1.05 m/s

(we could have arrived to the same result subtracting both sides in (1), and (2))

3 0

2 years ago

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0

2 years ago

A spaceship maneuvering near planet zeta is located at r⃗ =(600i^−400j^+200k^)×103km, relative to the planet, and traveling at v

slava [35]

<u>Answer:</u>

The spaceship's position when the engine shuts off = $(708.15 i - 444.1 j + 200 k)*10^3km$

<u>Explanation:</u>

Initial location of spaceship = (600 i - 400 j + 200 k)* $10^3km$ = (600 i - 400 j + 200 k)* $10^6m$

Initial velocity = 9500 i m/s

Acceleration = (40 i - 20 k) $10^3m/s^2$

Time = 35 minute = 35 * 60 = 2100 seconds

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Substituting

$s= (9500 i)*2100+\frac{1}{2}*(40 i - 20 k)*2100^2\\ \\ s=9500*2100 i+20*2100^2i-10*2100^2j\\ \\ s=19.95*10^6i+88.2*10^6i-44.1*10^6j\\ \\ \\s=(108.15i-44.1j)*10^6m$

So final position = $((600 i - 400 j + 200 k)+(108.15i-44.1j))*10^6=(708.15 i - 444.1 j + 200 k)*10^6m$

= $(708.15 i - 444.1 j + 200 k)*10^3km$

The spaceship's position when the engine shuts off = $(708.15 i - 444.1 j + 200 k)*10^3km$

3 0

2 years ago

Read 2 more answers