Does an electrochemical cell use two terminals or a voltmeter?

The pilot of an aircraft wishes to fly due west in a 33.9 km/h wind blowing toward the south. The speed of the aircraft in the a

diamong [38]

Answer: $\theta =9.96^{\circ}$ North of west

Explanation:

Given

Plane wishes to fly in west

but wind with speed 33.9 km/h towards south obstructing its path

so plane must fly at an angle of \theta w.r.t west such that it final velocity is towards west

Plane absolute speed=195 km/h

To fly towards west velocity in Y direction should be zero

thus $195sin\theta =33.9$

$\theta =9.96^{\circ}$

so Plane should head towards $9.96^{\circ}$ North of west in order to fly in west.

So plane

actual velocity is

$v=-195cos9.96\hat{i}+195sin9.96\hat{j}$

5 0

3 years ago

A 2,700-kg truck runs into the rear of a 1,000-kg car that was stationary. The truck and car are locked together after the colli

Leto [7]

Answer:

6200 J

Explanation:

Momentum is conserved.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

The car is initially stationary. The truck and car stick together after the collision, so they have the same final velocity. Therefore:

m₁ u₁ = (m₁ + m₂) v

Solving for the truck's initial velocity:

(2700 kg) u = (2700 kg + 1000 kg) (3 m/s)

u = 4.11 m/s

The change in kinetic energy is therefore:

ΔKE = ½ (m₁ + m₂) v² − ½ m₁ u²

ΔKE = ½ (2700 kg + 1000 kg) (3 m/s)² − ½ (2700 kg) (4.11 m/s)²

ΔKE = -6200 J

6200 J of kinetic energy is "lost".

3 0

3 years ago

Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur

BARSIC [14]

Answer:

$v_A=7667m/s\\\\v_B=7487m/s$

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

$F_g=\frac{GMm}{R^{2} }$

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

$F_c=m\frac{v^{2}}{R}$

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

$\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}$

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So $R_A=6774km=6.774*10^6m$ and $R_B=7103km=7.103*10^6m$ (Since $R_{earth}=6371km$ ). Then, we get:

$v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s$

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0

2 years ago

Which statement describes how work and power are similar?

postnew [5]

The statement that describes how work and power are similar is D. you must know time and energy to calculate both.
I am not completely sure though, so I hope this helps. :)

6 0

3 years ago

Read 2 more answers

A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. the hoist weighs 430 n. the ropes,

ICE Princess25 [194]

Refer to the diagram shown below.

For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)

For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)

Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N

Answer:
T₂ = 339.06 N
T₃ = 276.57 N

7 0

2 years ago