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ycow [4]
3 years ago
12

Edward travels 150 kilometers due west and then 200 kilometers in a direction 60° north of west. What is his displacement in the

westerly direction?
Physics
1 answer:
lions [1.4K]3 years ago
8 0
     We must decompose the displacement vector to find the new displacement to the west.

d_{x}=v.cos\O \\ d_{x}=200. \frac{1}{2}  \\ d_{x}=100km
  
     Adding all displacements to West, It comes:

\Delta S= d_{o}+d_{x} \\ \Delta S= 150+100 \\ \boxed {\Delta S= 250 km}

If you notice any mistake in English, please let me know, because I'm not native.

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A horizontal desk surface measures 1.7 m by 1.0 m. If the Earth's magnetic field has magnitude 0.42 mT and is directed 68° below
AlexFokin [52]

Answer:

The magnetic flux through the desk surface is 6.6\times10^{-4}\ T-m^2.

Explanation:

Given that,

Magnetic field B = 0.42 T

Angle =68°

We need to calculate the magnetic flux

\phi=BA\costheta

Where, B = magnetic field

A = area

Put the value into the formula

\phi=0.42\times10^{-3}\times1.7\times1.0\cos22^{\circ}

\phi=0.42\times10^{-3}\times1.7\times1.0\times0.927

\phi=6.6\times10^{-4}\ T-m^2

Hence, The magnetic flux through the desk surface is 6.6\times10^{-4}\ T-m^2.

3 0
3 years ago
Calculate the amount of momentum of an object with a mass of 10 kg travelling al a
irina [24]

Answer:

50kg.m/s

Explanation:

In order to find momentum you must use the formula P=mv

p= momentum

m=mass

v= velocity

so in other words, momentum= mass times velocity

or in this case, momentum= 10 times 5  :)

7 0
3 years ago
Help asap please I will give you 5stars
boyakko [2]

Explanation:

In the parallel combination, the equivalent resistance is given by :

\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+....

4. When three 150 ohms resistors are connected in parallel, the equivalent is given by :

\dfrac{1}{R}=\dfrac{1}{150}+\dfrac{1}{150}+\dfrac{1}{150}\\\\R=50\ \Omega

5. Three resistors of 20 ohms, 40 ohms and 100 ohms are connected in parallel, So,

\dfrac{1}{R}=\dfrac{1}{20}+\dfrac{1}{40}+\dfrac{1}{100}\\\\=11.76\ \Omega

Hence, this is the required solution.

6 0
3 years ago
You have a spring-loaded air rifle. When it is loaded, the spring is compressed 0.3 m and has a spring constant of 150 N/m. In j
Feliz [49]

The potential energy of the spring is 6.75 J

The elastic potential energy stored in the spring is given by the equation:

E= \frac{1}{2} kx^2

where;

k is the spring constant

x is the compression/stretching of the string

In this problem, we have the spring as follows:

k = 150 N/m is the spring constant

x = 0.3 m is the compression

Substituting in the equation, we get

E=\frac{1}{2} (150) (0.3)^2

E=6.75J

Therefore. the elastic potential energy stored in the spring is 6.75J .

Learn more about potential energy here:

brainly.com/question/10770261

#SPJ4

6 0
1 year ago
Will give Brainliest!! Question attached.
ss7ja [257]

Answer:

im pretty sure it is 3.0 K

Explanation:

8 0
3 years ago
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