Two forces of 50 N and 30N respectively, are acting on an object. Find the net force (in N) on the object if

Link walks 5 miles east, then 7 miles north then 2 miles south

If you draw a simple diagram you’ll see that your displacement from the start is 5 miles East and 7 miles North and then 2 miles south

These distances are the sides of a right-angled triangle. The displacement you are looking for is the hypotenuse of the triangle. Use the Pythagorean theorem to calculate it.

Displacement

= √ (5+2)^2 + 7^2)

= 9.899 miles

Distance

= 5+7+2

= 14 miles

Time to complete his whole journey = 4 hours

Average Speed= Distance travelled / Time

= 14÷4

= 3.5 miles/hour

Average velocity= Displacement / Time

= 9.899÷4

= 2.47475 miles/hour

The distance is 14 miles and the displacement is 9.899 miles, the Average Speed is 3.5 miles/hour and the Average Velocity is 2.47475 miles/hour

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A bowling ball results in kinetic friction compared to a sliding bowling ball.

According to Conservation of momentum, if a bowling ball hits some pins, the momentum that lost by the bowling ball is known to be equal to the momentum obtained by the pins.

The friction that is used in bowling is kinetic friction because the more oil that is placed down, the lower the friction that is found between the ball and that of the lane surface. The  little friction, the stronger it is for the bowler to be able to send the ball in a curved path and thus the formula to find the kinetic friction is know to be : µk=F k/mg.

Therefore, A bowling ball results in kinetic friction compared to a sliding bowling ball.

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The electric force on one of the masses is 0.6 N.

The acceleration of the mass is 0.35 m/s².

The electric force between the masses is calculated as follows;

F = kq²/r²

where;

F = (9 x 10⁹ x (9.8 x 10⁻⁶)²)/(1.2²)

F = 0.6 N

The acceleration of the mass is calculated as follows;

F = ma

a = F/m

a = (0.6 N) / (1.7 kg)

a = 0.35 m/s²

Thus, the electric force on one of the masses is 0.6 N.

The acceleration of the mass is 0.35 m/s².

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You throw a rock from the upper edge of a 75.0 -m vertical dam with a speed of 25.0 m/s at 65.0∘ above the horizon.

The horizon is the line that, when seen from a position on or near the surface of a celestial body, appears to separate the surface from the sky of that body. All viewing directions are split according to whether it crosses the surface of the relevant body or not.

Since the true horizon is an imaginary line, it can only be seen with any degree of accuracy when it is situated along a generally flat surface, such as the oceans of the Earth. On Earth, the geography may also cause biological objects like trees and/or man-made objects like buildings to obstruct this line in some locations. The location where these obstructions overlap the sky is known as the visible horizon.

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The required voltage is 10 Volt for 100 ohms of resistance and 0.1 ampere of current.

electric current is defined as the rate at which charge passes a fixed unit cross sectional area.

voltage is defined as the physical quantity required to move charge from one end to other.

Resistance is defined as the the factor opposing electric current.

Ohms law states that voltage is directly proportional to electric current

V=IR

given:

current (I) = 0.1 A

resistance (R)=100 ohms

now from definition of voltage we have:

V=I.R

V=100×0.1

V=10volts

therefore the required voltage is 10 volts.

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The car travels 25 meters before coming to rest and will hit the deer.

Given in the Question,

Initial speed = u = 15 m/s

Deceleration = a = 4.5 m/s²

According to the question, if the car stops before traveling 20m, it will avoid hitting the deer. So, we need to find the stopping distance for the car.

Deceleration is negative acceleration, so the sign of acceleration is will be -.

Also, the car comes to rest after applying the brakes. Therefore the final speed of the car will be zero.

Final speed = v = 0 m/s

Using the Third equation of motion,

v² - u² = 2as

Put in the values, we get

(0)² - (15)² = 2(-4.5)s

-2×4.5 × s = - 15× 15

s = 225/9

s = 25 m

So the car will come to rest after traveling 25 meters. But the Deer is present at 20 meters; therefore the car will hit the deer.

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The magnitude of the minimum deceleration needed to avoid the accident is –0.89 m/s²

The solution of the question can be obtained by first approaching the relation between Speed, Distance and Time which is given by:

Speed = Distance/Time

Now according to question;

Speed = 18 m/s

Time = 0.54 s

Therefore, Distance = Speed×Time

Distance = 18 × 0.54

Distance = 9.72 m

Thus, the engineer travelled a distance of 9.72 m during the reaction time.

Now, there is a need to find distance between the engineer and the car. This can be obtained by:

Distance between the engineer and the car = Total distance - Distance during the reaction time

Distance between the engineer and the car = 190 – 9.72

Distance between the engineer and the car = 180.28 m

Finally, we need to determine the magnitude of the deceleration needed to avoid the accident. This can be obtained by using newton's third equation of motion, which is represented as:

v² = u² + 2as; where u is the initial velocity, v is the final velocity, a is the acceleration/deceleration and S is the distance.

According to the question;

Initial velocity (u) = 18 m/s

Final velocity (v) = 0 m/s

Distance (s) = 180.28 m

Deceleration (a) =?

Thus, using Newton's third equation of motion

0² = 18² + (2 × a × 180.28)

0 = 324 + 360.56a

=> –324 = 360.56a

=> a = –324 / 360.56

=> a = –0.89 m/s²

Therefore, the magnitude of the deceleration needed to avoid the accident is –0.89 m/s.

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Explanation:

probably luck or the right material's

The attractive force of Earth on the moon will be said to be double when there is a doubling of the mass of the moon.

Since the moon has an attracting force, it is one that will remain the same on Earth. In gravitational forces, the two concerned objects always feel the same force.

The explanation of how this would affect the gravity between earth and the moon is that this can result to the earth tilt a little bit harder to change, which could imply that there will be a more stable climate and  ice ages may not occur as often.

Therefore, The attractive force of Earth on the moon will be said to be double when there is a doubling of the mass of the moon. Since the moon has an attracting force, it is one that will remain the same on Earth. In gravitational forces, the two concerned objects always feel the same force.

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Answer:

13.33 mA s

Explanation:

400 mA * 1/30 s = 13.33 mA s

The angular momentum can be indicated by formula mvr. The correct option is A.

The property of any rotating object given by moment of inertia times angular velocity is defined as angular momentum.

It is the property of a rotating body determined by the product of the rotating object's moment of inertia and angular velocity.

The rotational equivalent of linear momentum is the angular momentum formula. Both concepts are concerned with the rate at which anything moves.

It can be represented by symbol L.

L = mvr.

Thus, the correct option is A.

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Answer: 1 > 3 > 2

Explanation: The range will increase with the velocity. If they are all launched at the same time the ones that are launched the hardest or with the most velocity will go the furthest horizontally. But because they are all launched from the same height and the force of gravity on all three projectiles is constant (the same for all three) they will all hit the ground at the same time but different distances from the starting point.

Answer:

35

Explanation:

It will take the fugitive 4.83 s to catch the empty box car

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

The time taken for the fugitive to catch the car if he maintains his maximum speed can be obtained as follow:

a = (v – u) / t

Thus,

t = (v – u) / a

t = (5.8 – 0) / 1.2

t = 5.8 / 1.2

t = 4.83 s

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The magnitude of the change in velocity is determined as 408.94 km/h.

The change in direction of the velocity is 64.4⁰ north of west.

The magnitude of the average acceleration during the trip is 0.0105 m/s².

The magnitude of change in velocity is the resultant velocity of the plane.

v² = a² + b² - 2ab cosθ

where;

v² = (192²) + (250²) - 2(192 x 250) cos(135)

v² = 167,236

v = √167,236

v = 408.94 km/h

vyi = 192 km/h

vy2 = 250 x sin(45) = 176.77 km/h

vy(total) = 192 km/h + 176.77 km/h = 368.77 km/h

vxi = 0

vx2 = - 250 km/h x cos(45) = -176.77 km/h

θ = arc tan (Vy/Vx)

θ = arc tan(368.77 / -176.77)

θ = -64.4 ⁰

θ = 64.4⁰ north of west.

a = v/t

v = 408.94 km/h = 113.6 m/s

h = 3 h = 10,800 seconds

a = (113.6 m/s) / ( 10,800 s)

a = 0.0105 m/s²

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The impulse the gravity imparts on the stone from the instant it was dropped until it hits the ground is 72.66 Ns

This is defined as the change in momentum of an object.

Impulse = change in momentum

Impulse = final moment – Initial momentum

Impulse = force × time

h = ½gt²

6.02 = ½ × 9.8 × t²

6.02 = 4.9 × t²

Divide both side by 4.9

t² = 6.02 / 4.9

Take the square root of both side

t = √(6.02 / 4.9)

t = 1.1 s

Impulse = Force × time

Impulse = 66.052 × 1.1

Impulse = 72.66 Ns

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The time of flight of the ball is 17.12 seconds.

The horizontal distance or range of the ball is 354.8 m.

The time of flight of the ball is calculated as follows;

T = (2u sinθ)/g

where;

T = (2 x 25 sin34) / (¹/₆ x 9.8)

T = 17.12 s

The range of the golf ball is calculated as follows;

R = Uxt

R = (U cosθ)t

R = (25 cos34) x 17.12

R = 354.8 m

Thus, the time of flight of the ball is 17.12 seconds.

The horizontal distance or range of the ball is 354.8 m.

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The correct answer is :

Here 100 N force is applied to make the box move with constant velocity from rest. That means 100 N force is applied to overcome the limiting  static friction and as soon as 100 N force is applied it starts moving.

Now,

Constant velocity means acceleration = 0

Net force acting on the box =mass × accelaration = mass × 0 = 0

Conceptually it is zero as it is balanced by kinetic friction which has equal value that of applied force. Because net force =Applied force - friction force and hence here friction force =applied force.

If there was any accelaration then there would exist a net force and then frictional force and applied force will be the same.  

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The space cruiser is 11070000 m far from the planet's surface

The following data were obtained from the question:

We can obtain the distance of the space cruiser from the planet's surface as follow:

s = ut + ½gt²

s = (0 × 600) + (½ × 61.5 × 600²)

s = 0 + (½ × 61.5 × 360000)

s = 0 + 11070000

s = 11070000 m

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The speed over the ground is 0 m/s.

The speed of an object or body is the magnitude of the change of its position over time or the magnitude of the change of its position per unit of time; it is thus a scalar quantity.

As it is given that the mosquito is flying at 2 m/s relatives to the air caught in a 2 m/s right angle crosswind.

As we know that speed over the ground is the difference between the flight speed and resistance speed.

So, the general equation for the speed over the ground is :

v = Flight Speed of the mosquito - Resistance Speed of the crosswind

This implies, [tex]v= 2 {~}m / s - 2{~} m / s[/tex]

[tex]v = 0 {~}m / s[/tex]

Hence, the speed over the ground is 0 m/s.

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Answer:

100

Explanation:

3-2=1

100/1=100

The statement that best explains why this helps protect orange trees and fruit from freezing is that the transfer of thermal energy in water as it freezes protects the orange trees by transferring heat energy from the orange trees and fruit. That is option A.

Freezing is the phase transition that involves the change of a liquid substance into a solid form as it's temperature becomes lower.

During winter seasons which is a weather that can freez the orange fruits, the orange grower sprays water on their orange trees to prevent it from freezing.

The mechanism that prevents the orange from freezing is the transfer of the thermal energy between the water and the orange fruits.

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The kinds of materials can be involved in static electric effects is All types of materials, both metals and non metals.

The static electric effects can be described as the effect that bring about the  changes to the  body  as a result of the changes in the distribution that is been imposed on the electric charges  that is found on the surface of the body.

It should be noted that sufficiently large surface charge density  could be discovered as a result of the effect, however in the case above option A is correct.

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(a) The acceleration is 0.2 m /s²

(b) The acceleration is  - 0.2 m /s²

(c) The acceleration is  - 0.3 m /s²

(d) The acceleration is 0.4 m /s²

(a)  Putting  v₁ₓ = 0.8 m/s , v₂ₓ = 1.2 m/s ,  t₁ = 2 s and t₂ = 4s  in equation (1)

, we get   a = 1.2 - 0.8 / 4 -2

              a = 0.4 / 2

               a = 0.2 m /s²

(b) Putting v₁ₓ = 1.6 m/s, v₂ₓ = 1.2 m/s , t₁ = 2 s and t₂ = 4s  in equation (1) , we get   a = 1.2 - 1.6 / 4 -2

              a = - 0.4 / 2

               a = - 0.2 m /s²

(c) Putting  v₁ₓ = - 0.4 m/s, v₂ₓ = - 1.0 m/s,  t₁ = 2 s and t₂ = 4s  in equation (1) , we get   a = -1.0 - (-0.4) / 4 -2

              a = - 0.6 / 2

               a = - 0.3 m /s²

(d) Putting  v₁ₓ = -1.6 m/s, v₂ₓ = -0.8 m/s,  t₁ = 2 s and t₂ = 4s  in equation (1) , we get   a = -0.8 - (-1.6) / 4 -2

              a = 0.8 / 2

               a = 0.4 m /s²

The traveller races in cases (a) and (c) and slows down in (b) and (d), however the common acceleration is positive in (a) and (d) and negative in (b) and (c). In alternative words, negative acceleration doesn't essentially indicate a deceleration down.

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The complete question is given below :

An astronaut has left the space shuttle on a tether to test a new personal maneuvering device. She moves along a straight line directly away from the shuttle. Her onboard partner measures her velocity before and after certain maneuvers, and obtains the following results:

(a) v₁ₓ = 0.8 m/s, v₂ₓ = 1.2 m/s; (speeding up)

(b)v₁ₓ = 1.6 m/s, v₂ₓ = 1.2 m/s; (slowing down)

(c) v₁ₓ = - 0.4 m/s, v₂ₓ = - 1.0 m/s; (speeding up)

(d) v₁ₓ = -1.6 m/s, v₂ₓ = -0.8 m/s; (slowing down)

If t₁ = 2 s and t₂ = 4s in each case, find the average acceleration for each set of data.

First, see that in the time interval 3.44 to 9.49, the average velocity is [tex]2.9\text{m} / (9.49\text{s} - 3.44\text{s})=2.9\text{m}/6.05\text{s} \approx 0.479\text{m/s}[/tex]. So, as we have uniform acceleration, the velocity must be linearly increasing over this entire interval, so for the average to be 0.479 m/s over this interval, the velocity must be 0.479 m/s in the exact middle of this interval, or at 5.465s.

We now note that the object starts from rest, which means that at 0s, the velocity is 0 m/s.  So, in 5.465 seconds, the velocity increases by 0.479 m/s.  We again have that the object undergoes uniform acceleration, meaning that the acceleration over this interval is a constant [tex]\frac{0.479\text{m/s}}{5.465\text{s}} \approx 0.0876 \text{m/s}^2[/tex].

Finally, note again that as we are looking at uniform acceleration, by the same principle at the beginning, the average velocity of the object during the time interval from 18.6s to 22.92s is the same as the velocity at the exact middle of this interval, or at 20.76s.  We have that acceleration is constant and 0.0876 m/s^2, and initial velocity is 0 at 0s. So, in 20.76 seconds, the object will have accelerated [tex]0.0876\text{m/s}^2 \cdot 20.76\text{s} \approx 1.82 \text{m/s}[/tex].

So, average velocity will be 1.82 m/s over the time interval 18.6s to 22.92s.