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Yanka [14]
2 years ago
10

a plane travels 204 km, northeast in 15.0 minutes. It also increases elevation by 1.6 km, upward in the same amount of time. Wha

t are the average velocities in the two separate directions?
Physics
1 answer:
mina [271]2 years ago
7 0

Answer:

230 m/s northeast, 1.8 m/s up

Explanation:

204 kilometres = 204000 metres

15.0 minutes = 900 seconds

Velocity = Distance / Time

= 204000 / 900

= 230 m/s northeast (to 2 sf.)

1.6km = 1600 metres

Velocity = 1600 / 900

= 1.8 m/s up (to 2 sf.)

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14250. I just took it

4 0
2 years ago
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Bill leaves his 60 W desk lamp on every day, including weekends, for eight hours. After one month (30 days), how much total ener
maxonik [38]

' W ' is the symbol for 'Watt' ... the unit of power equal to 1 joule/second.

That's all the physics we need to know to answer this question.
The rest is just arithmetic.

(60 joules/sec) · (30 days) · (8 hours/day) · (3600 sec/hour)

= (60 · 30 · 8 · 3600) (joule · day · hour · sec) / (sec · day · hour)

= 51,840,000 joules
__________________________________

Wait a minute !  Hold up !  Hee haw !  Whoa ! 
Excuse me.  That will never do.
I see they want the answer in units of kilowatt-hours (kWh).
In that case, it's

(60 watts) · (30 days) · (8 hours/day) · (1 kW/1,000 watts)

= (60 · 30 · 8 · 1 / 1,000) (watt · day · hour · kW / day · watt)

= 14.4 kW·hour

Rounded to the nearest whole number:

14 kWh

7 0
2 years ago
A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward
AleksandrR [38]

Answer:

(a)  104 N

(b) 52 N

Explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.  

The component of F perpendicular to the ramp is Fy.

(a)  

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.  

Resolve F into its x-component from Pythagorean theorem:  

Fx=Fcos30°

Solve for F:  

F= Fx/cos30°  

Substitute for Fx from given data:  

Fx=90 N/cos30°

   =104 N

(b) Resolve r into its y-component from Pythagorean theorem:

     Fy = Fsin 30°

   Substitute for F from part (a):

     Fy = (104 N) (sin 30°)  

          = 52 N  

5 0
2 years ago
A truck collides with a car on horizontal ground. At one moment during the collision, the magnitude of the acceleration of the t
Mice21 [21]

Answer:

The magnitude of the acceleration of the car is 35.53 m/s²

Explanation:

Given;

acceleration of the truck, a_t = 12.7 m/s²

mass of the truck, m_t = 2490 kg

mass of the car, m_c = 890 kg

let the acceleration of the car at the moment they collided = a_c

Apply Newton's third law of motion;

Magnitude of force exerted by the truck = Magnitude of force exerted by the car.

The force exerted by the car occurs in the opposite direction.

F_c = -F_t\\\\m_ca_c = -m_t a_t\\\\a_c =- \frac{m_ta_t}{m_c} \\\\a_c = -\frac{2490 \times 12.7}{890} \\\\a_c = - 35.53 \ m/s^2

Therefore, the magnitude of the acceleration of the car is 35.53 m/s²

3 0
2 years ago
An atom of argon has a radius of 71.pm and the average orbital speed of the electrons in it is about ×3.9107/ms. calculate the l
Anna11 [10]

Answer: 2.1 %

Explanation:

The radius of the Argon atom, r = 71 pm = 7.1 × 10 ⁻¹¹ m

Average orbital speed of electrons, v = 3.9 × 10⁷ m/s

From uncertainty principle:

Δx m Δv ≥ h/4π

mass of electron, m = 9.1 ×10⁻³¹ kg

Δx = radius of the argon atom = 7.1 × 10 ⁻¹¹ m

\Rightarrow \Delta v = \frac {6.626 \times 10^{-34} m^2kg/s}{4\times 3.14 \times 7.1 \times 10^{-11} m \times 9.1 \times 10^{-31} kg}

\Delta v = 8.2 \times 10^5 m/s

Percentage uncertainty:

\frac{\Delta v}{v} \times 100\% = \frac {8.2 \times 10^5 m/s}{3.9 \times 10^7 m/s} \times 100 \%= 2.1 \%

7 0
2 years ago
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