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Leni [432]
2 years ago
15

un avión se mueve en línea recta a una velocidad constante de 400 km/h durante 1,5h de su recorrido ¿que distancia recorrió en e

se tiempo?
Physics
1 answer:
Amanda [17]2 years ago
6 0

The distance covered by the plane is 600 km

Explanation:

The motion of the plane is a uniform motion, so at constant velocity, therefore we can use the following equation

d=vt

where

d is the distance covered

v is the velocity

t is the time interval considered

For the plane in this problem,

v = 400 km/h

t = 1.5 h

Substituting, we find the distance covered:

d=(400)(1.5)=600 km

Learn more about distance:

brainly.com/question/3969582

#LearnwithBrainly

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11. A vector M is 15.0 cm long and makes an angle of 20° CCW from x axis and another vector N is 8.0 cm long and makes an angle
Alja [10]

Answer:

The magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°

Explanation:

To find the resultant vector, you first calculate x and y components of the two vectors M and N. The components of the vectors are calculated by using cos and sin function.

For M vector you obtain:

M=M_x\hat{i}+M_y\hat{j}\\\\M=15.0cm\ cos(20\°)\hat{i}+15.0cm\ sin(20\°)\hat{j}\\\\M=14.09cm\ \hat{i}+5.13\ \hat{j}

For N vector:

N=N_x\hat{i}+N_y\hat{j}\\\\N=8.0cm\ cos(40\°)\hat{i}+8.0cm\ sin(40\°)\hat{j}\\\\N=6.12cm\ \hat{i}+5.142\ \hat{j}

The resultant vector is the sum of the components of M and N:

F=(M_x+N_x)\hat{i}+(M_y+N_y)\hat{j}\\\\F=(14.09+6.12)cm\ \hat{i}+(5.13+5.142)cm\ \hat{j}\\\\F=20.21cm\ \hat{i}+10.27cm\ \hat{j}

The magnitude of the resultant vector is:

|F|=\sqrt{(20.21)^2+(10.27)^2}cm=22.66cm

And the direction of the vector is:

\theta=tan^{-1}(\frac{10.27}{20.21})=29.93\°

hence, the magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°

4 0
3 years ago
To get copper from the solid phase to the liquid phase must of the copper sample
Elina [12.6K]
You would want to increase the temperature. 
8 0
2 years ago
Read 2 more answers
A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coe
andrey2020 [161]

Answer:

The magnitude of force is 1593.4N

Explanation:

The sum of the horizontal components of the friction and the normal force will be equal to the centripetal force on the car. This can be represented as

fcostheta + Nsintheta = mv^2/r

Where F = force of friction

Theta = angle of banking

N = normal force

m = mass of car

v = velocity of car

r = radius of curve

The car has no motion in the vertical direction so the sum of forces = 0

The vertical component of the normal force acts upwards whereas the weight of the car and the vertical component friction acts downwards.

Taking the upward direction to be positive,rewrite the equation above to get:

Ncos thetha = mg - fsintheta =0

Ncistheta = mg + fain theta

N = mg/cos theta + sintheta/ costheta

fcostheta +[mg/costheta + ftan theta] sin theta = mv^2/r

Substituting gives:

f = (1/(costheta + tanthetasintheta) + mgtantheta = mv^2/r - mgtantheta)

Substituting given values into the above equation

f = 1/(cos25 + tan 25 )(sin25)[ 600×30/120 - (600×9.81)tan

f = 1593.4N25

4 0
3 years ago
Read 2 more answers
Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 21.0 cm and carries a
Burka [1]

Solution :

a). B at the center :

     $=\frac{u\times I}{2R}$

Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.

Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE

b). Also, the sum of the fields must be zero.

Therefore,

$\left(\frac{u\times I_1}{2R_1}\right) + \left(\frac{u\times I_2}{2R_2}\right) = 0$

So,

$\frac{I_1}{d_1}= \frac{I_2}{d_2}$

$=\frac{16}{21}=\frac{I_2}{32}$

$I_2=24.38 $ A

Therefore, the current in the outer wire is 24.38 ampere.

3 0
2 years ago
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a cylinder has a radius of 2.1cm and a length of 8.8cm .total charge 6.1 x 10^-7C is distributed uniformly throughout. the volum
7nadin3 [17]

The answer for the following problem is explained below.

Therefore the volume charge density of a substance (ρ) is 0.04 × 10^{-1} C.

Explanation:

Given:

radius (r) =2.1 cm = 2.1 ×10^{-2} m

height (h) =8.8 cm = 8.8 ×10^{-2} m

total charge (q) =6.1×10^{-7} C

To solve:

volume charge density (ρ)

We know;

<u> ρ =q ÷ v</u>

volume of cylinder = π ×r  × r × h

volume of cylinder =3.14 × 2.1 × 2.1 ×10^{-4} × 8.8 ×10^{-2}

volume of cylinder (v) = 122.23 ×10^{-6}

<u> ρ =q ÷ v</u>

ρ = 6.1×10^{-7} ÷   122.23 ×10^{-6}

<u>ρ = 0.04 × </u>10^{-1}<u> C</u>

Therefore the volume charge density of a substance (ρ) is 0.04 × 10^{-1} C.

3 0
3 years ago
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