Answer:
I assume the graph is looking like in the picture bellow.

North component:

cos(41.5) * 835 = 625.37 km/h

West component of speed:

sin(41.5) * 835 = 553.29 km/h

After 2.2 hours plane will fly:

2.2*625.37 = 1375.81 km north

2.2*553.29 = 1217.23 km west

North component:

cos(41.5) * 835 = 625.37 km/h

West component of speed:

sin(41.5) * 835 = 553.29 km/h

After 2.2 hours plane will fly:

2.2*625.37 = 1375.81 km north

2.2*553.29 = 1217.23 km west

Answer:
### Final answer:

### Explanation:

### Learn more about Velocity components here:

To find the components of the velocity vector, you can use **trigonometry**. The north component is calculated using the sine function and the west component is calculated using the cosine function. After 2.20 hours, the distance traveled north and west can be found by multiplying the velocity components by the time.

To find the components of the velocity vector in the northerly and westerly directions, we can use trigonometry. The velocity vector is 835 km/h and is traveling in a direction 41.5° west of north. To find the north component, we can use the sine function: **North component = velocity * sin(angle)**. To find the west component, we can use the cosine function: **West component = velocity * cos(angle)**.

After 2.20 hours, we can find the distance traveled north and west by multiplying the velocity components by the time: **Distance north = North component * time** and **Distance west = West component * time**.

Let's calculate the values:

- North component = 835 km/h * sin(41.5°)
- West component = 835 km/h * cos(41.5°)
- Distance north = North component * 2.20 h
- Distance west = West component * 2.20 h

#SPJ3

The electric field at the distance of 3.5 meters from an infinite wall of charges is 125 N/C. What is the magnitude of the electric field 1.5 meters from the wall?

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used instead for alpha particles with energy 300 keV . An alpha particle has charge q=+2e and massm=6.64×10−27kg .If the magnetic field isnt changed, what will be the orbital radius of the alpha particles? Express your answer to three significant figures and include the appropriate units.

On December 26, 2004, a great earthquake occurred off the coast of Sumatra and triggered immense waves (tsunami) that killed some 200000 people. Satellites observing these waves from space measured 800 from one wave crest to the next and a period between waves of 1.0 hour.Part AWhat was the speed of these waves in m/s?Express your answer using two significant figures.=Part BWhat was the speed of these waves in km/h ?Express your answer using two significant figures.=

When a body falls freely under gravity, then the work done by the gravity is ___________

A car (m = 2000 kg) is going around an unbanked curve at the recommended speed of 11m/s (24.6 MPH). (a) If the radius of the curvature of the path is 25m and the coefficient of static friction between the rubber tires and the road is µs = 0.70, does the car skid as it goes around the curve? (b) What will happen if the driver ignores the highway speed limit sign and travels at 18 m/s (40.3 MPH)? (c) What speed is safe for traveling around the curve if the road surface is wet from a recent rainstorm and the coefficient of static friction between the wet roud and the rubber tires is µs = 0.50?

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used instead for alpha particles with energy 300 keV . An alpha particle has charge q=+2e and massm=6.64×10−27kg .If the magnetic field isnt changed, what will be the orbital radius of the alpha particles? Express your answer to three significant figures and include the appropriate units.

On December 26, 2004, a great earthquake occurred off the coast of Sumatra and triggered immense waves (tsunami) that killed some 200000 people. Satellites observing these waves from space measured 800 from one wave crest to the next and a period between waves of 1.0 hour.Part AWhat was the speed of these waves in m/s?Express your answer using two significant figures.=Part BWhat was the speed of these waves in km/h ?Express your answer using two significant figures.=

When a body falls freely under gravity, then the work done by the gravity is ___________

A car (m = 2000 kg) is going around an unbanked curve at the recommended speed of 11m/s (24.6 MPH). (a) If the radius of the curvature of the path is 25m and the coefficient of static friction between the rubber tires and the road is µs = 0.70, does the car skid as it goes around the curve? (b) What will happen if the driver ignores the highway speed limit sign and travels at 18 m/s (40.3 MPH)? (c) What speed is safe for traveling around the curve if the road surface is wet from a recent rainstorm and the coefficient of static friction between the wet roud and the rubber tires is µs = 0.50?

**A****.**The **time taken** for the **car** to **stop** is **8.75 s**

**B****.**The **distance travelled **when the **brakes** were **applied** till the **car stops** is **136.89 m**

**A**. Determination of the **time taken** for the **car** to **stop**.

**We'll**begin**by****calculating**the**deceleration**of the**car**

Initial velocity (u) = 70 mph = 0.447 × 70 = 31.29 m/s

Final velocity (v) = 30 mph = 0.447 × 30 = 13.41 m/s

Time (t) = 5 s

**Finally****,**we shall**determine**the**time taken**for the**car**to**stop**.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Thus, the **time taken** for the **car** to **stop** is **8.75 s**

**B****.**Determination of the **total distance travelled** when the **brakes** were **applied**.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Therefore, the **total distance travelled **by the **car** when the **brakes** were **applied** is **136.89 m**

Learn more: brainly.com/question/9163788

**Answer:8.75 s,**

**136.89 m**

**Explanation:**

Given

Initial velocity

velocity after 5 s is

Therefore acceleration during these 5 s

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

(b)total distance traveled before stoppage

s=136.89 m

**Answer:**

**475 N/C**

**Explanation:**

As we know that, the **electric field in parallel plate capacitor is same (constant) throughout**. And is potential gradient.

So, Electric field is given by

**Electric field = potential gradient**

Here, the potential change is 3.8V and the distance from negative plate to positive plate is 1.6 cm. **The potential from negative plate to the center is (1.6/2)cm i.e., 0.8 cm.**

But we have to take **distance in SI units** So, distance=

So, Electric field is

So, **electric field is 475 Volts per meter. **

Note : Also we can say **475 Newtons per coulomb**

**Answer:**

the googles are 5.3 m from the edge

**Explanation:**

**Given that**

depth of pool , d = 3.2 m

Now, let i be the angle of incidence

a laser pointer 0.90 m above the edge of the pool and laser beam enters the water 2.2 m from the edge

⇒tan i = 2.2/0.9

solving we get

i = 67.8°

Using snell's law ,

n1 ×sin(i) = n2 ×sin(r)

n1= refractive index of 1st medium= 1

n2= refractive index of 2nd medium = 1.33

r= angle of reflection

therefore,

r = 44.1°

Now,

distance of googles = 2.2 + d×tan(r)

distance of googles = 2.2 + 3.2×tan(44.1)

distance of googles = 5.3 m

the googles are 5.3 m from the edge

**Answer:**

V1 = -3.260 m/s, V2 = 1.303 m/s

**Explanation:**

Let mass of the left glider m1 = 0.157 kg and velocity v1 = 0.850 m/s

mass of the right glider m2 = 0.306 Kg and v2 = -2.26 m/s (-ve sign mean it is opposite to direction of left glider)

To Find: Final Velocity of Left Glider is V1=? m/s and Velocity of right Glider is V2 =? m/s (After Collision)

from law of conservation of momentum and energy we deduce a formula:

V1 = (m1-m2) v1 /(m1+m2) + 2 m2 v2/(m1+m2)

V1 = (0.157 kg - 0.306 Kg) × 0.850 m/s / (0.157 kg + 0.306 Kg) + 2 ×0.306 kg × -2.26 m/s / (0.157 kg + 0.306 Kg)

V1 = -0.273 -2.987

V1 = -3.260 m/s

and V2 Formula

V2 = (m2-m1) v2/(m1+m2) + 2 m1 v1/(m1+m2)

V2 = (0.157 kg - 0.306 Kg) × -2.26 m/s / (0.157 kg + 0.306 Kg) + 2 ×0.157 kg × 0.850 m/s / (0.157 kg + 0.306 Kg)

V2 = 0.727 + 0.576

V2 = 1.303 m/s

-0.149, 0.463

**Answer:**

a) Current = 11 mA

b) Energy = 66 mJ

c) Power = 101.54 W

**Explanation:**

a) Voltage, V = IR

Voltage, V = 6 V, Resistance, R = 550 Ω

Current, I

b) Energy = Current x Voltage = 6 x 0.011 = 0.066 J = 66 mJ

c)

b. the heavier object receives a larger impulse

c. both objects receive the same impulse

d. the answer depends on the ratio of the masses

e. the answer depends on the ratio of the speeds

**Answer:**

**Explanation:**

The impulse has as an expression

I = ∫ F dt

In addition, on impulse of isolated systems is

I =ΔP = m vf - m v₀

Let's replace

F dt = m (vf -vo)

The force in these shocks laqueunxcyerpoesjerce on the other, using the law of action and reaction state forces has the same magnitude and the time of the shock is equal for the two bodies, this implies that the impulse is equal for the two bodies

Let's check the answers

a) False as the forces are of action and reaction the impulse must be equal

b) False

c) True. Why do we have an action and reaction stop

d False

e) false

In collision between two objects, despite difference in their masses, both objects receive the same impulse, as per **Newton's third law.**

In a collision between two objects with unequal masses, according to Newton's third law of motion, **'For every action, there is an equal and opposite reaction'**, the magnitude of the impulse imparted to the lighter object by the heavy object is exactly the same as the magnitude of the impulse imparted to the heavier object by the lighter one.

This is regardless of the masses or speeds of the objects involved. The direction of these impulses will be opposite, but their magnitudes will be the equal. So, option c. 'both objects receive the same impulse' is correct.

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