As the angle between two concurrent forces is
decreased from 55° to 15°, the magnitude of
the resultant force

Answer:

B the slope of car postion time graph

Explanation:

Answer:

The slope of the car's position time graph.

Explanation:

Given that,

Force in right side = 50 N

Force in left side = 110 N

To find,

The resultant and direction of force.

Solution,

Let the net force F is acting on the rope. Also, we can assume that right side is positive and left side is negative.

F = 50+(-110)

= -60 N

So, the magnitude of the resultant force acting on the rope is 60 N and it acts in left side.

Answer:

It increases the time it takes for the person to stop.

Explanation:

Answer:

C: It increases the time it takes for the person to stop.

Explanation:

on edge! hope this helps!!~ o(〃＾▽＾〃)o

Given that,

A whale dives at an angle of 24 degrees to the horizontal surface of the water.

The whale continues in a straight line 145 m.

To find,

The horizontal and vertical components of the displacement of the whale.

Solution,

The horizontal component of displacement is :

[tex]d_x=d\cos\theta\\\\=145\times \cos(24)\\\\=132.46\ m[/tex]

The vertical component of displacement is :

[tex]d_y=d\sin\theta\\\\=145\times \sin(24)\\\\=58.97\ m[/tex]

Answer:

16294 rad/s

Explanation:

Given that

M(ns) = 2M(s), where

M(s) = 1.99*10^30 kg, so that

M(ns) = 3.98*10^30 kg

Again, R(ns) = 10 km

Using the law of gravitation, the force between the Neutron star and the sun is..

F = G.M(ns).M(s) / R²(ns), where

G = 6.67*10^-11, gravitational constant

Again, centripetal force of the neutron star is given as

F = M(ns).v² / R(ns)

Recall that v = wR(ns), so that

F = M(s).w².R(ns)

For a circular motion, it's been established that the centripetal force is equal to the gravitational force, hence

F = F

G.M(ns).M(s) / R²(ns) = M(s).w².R(ns)

Making W subject of formula, we have

w = √[{G.M(ns).M(s) / R²(ns)} / {M(s).R(ns)}]

w = √[{G.M(ns)} / {R³(ns)}]

w = √[(6.67*10^-11 * 3.98*10^30) / 10000³]

w = √[2.655*10^20 / 1*10^12]

w = √(2.655*10^8)

w = 16294 rad/s

Answer:

sure I will helpy you iru

Given that,

Mass of a sprinter = 70 kg

Initial velocity, u = 0

Final velocity, v = 10 m/s

Time, t = 3 s

To find,

Power output.

Solution,

The work done by the sprinter is equal to its kinetic energy.

[tex]W=\dfrac{1}{2}m(v^2-u^2)\\\\=\dfrac{1}{2}\times 70\times 10^2\\\\=3500\ J[/tex]

Let P is power output. Power is equal to work done per unit time. So,

[tex]P=\dfrac{3500\ J}{3\ s}\\\\=1166.67\ W[/tex]

So, the power output is 1166.66 W.

Answer:

132 N

Explanation:

Given that a  1.1 kg hammer strikes a nail. Before the impact, the hammer is moving at 4.5 m/s; after the impact it is moving at 1.5 m/s in the opposite direction. If the hammer is in contact with the nail for 0.025 s, what is the magnitude of the average force exerted by the hammer on the nail

From Newton 2nd law of motion,

Change in momentum = impulse.

Change in momentum = m( V - U )

Substitute all the parameters into the formula

Change in momentum = 1.1 ( 4.5 - 1.5 )

Change in momentum = 1.1 × 3

Change in momentum = 3.3 kgm/s

Impulse = Ft

That is,

Ft = 3.3

Substitute time t into the formula above

F × 0.025 = 3.3

F = 3.3 / 0.025

F = 132 N

Therefore, the magnitude of the average force exerted by the hammer on the nail is 132 N.

Answer:

c

Explanation:

Answer:

C

Explanation:

got it right on edge

Answer:

destructive interference occurs

A) the forces are acting in the same direction.. B) Together, forces are acting in opposite directions

Answer:

A) 80 N

B) 20 N

Explanation:

A) If the forces acting are in the same direction, then the net force will be a sum of both so many faces..

Thus;

ΣF = 50 + 30

ΣF = 80 N

B) If the forces are acting in the in opposite directions with the larger force pointing in the positive y-axis then, the net force is;

ΣF = 50 - 30

ΣF = 20 N

F=nmv

where;

n=no. of bullets = 1

m=mass of bullets=2g *10^-3

V=velocity of bullets200m/sec

F=1

loss in Kinetic energy=gain in heat energy

1/2MV^2=MS∆t

let M council M

=1/2V^2=S∆t

M=2g

K.E=MV^2/2

=(2*10^-3)(200)^2/2

2 councils 2

2*10^-3*4*10/2

K.E=40Js

H=mv∆t

(40/4.2)

40Js=40/4.2=mc∆t

40/4.2=2*0.03*∆t

=158.73°C

Answer:

[tex]Vf^2=Vo^2+2aS\\(0m/s)^2=(4.67m/s)^2+(2*-10m/s^2)S\\-(4.67)^2 m^2/s^2=-20m/s^2*S\\S=(21.8089/20) m\\S=1.090445 m\\[/tex]

Answer:

Explanation:

The path of the Earth–Moon system in its solar orbit is defined as the movement of this mutual centre of gravity around the Sun. Consequently, Earth's centre veers inside and outside the solar orbital path during each synodic month as the Moon moves in its orbit around the common centre of gravity.

Answer:

B

Explanation

the moon is constantly falling but moving fast enough to continue falling around the earth.

Answer:

5.14m/s

Explanation:

First we need to get the Maximum height reached by the rocket;

H = u²sin² theta/2g

H = 25²sin² 40/2(9.8)

H = 625(0.5878)/19.6

H = 18.74m

Get the velocity after 1.5secs

Using the equation of motion

H = ut + 1/2gt²

18.74 = u(1.5)+1/2(9.8)(1.5)²

18.74 = 1.5u + 4.9(2.25)

18.74 = 1.5u + 11.025

1.5u = 18.74 - 11.025

1.5u = 7.715

u = 7.715/1.5

u = 5.14m/s

Hence the velocity after 1.5secs is 5.14m/s

Answer:

Speed of the approaching train = 15.45 m/s

Explanation:

Given:

Frequency F0 = 220 Hz

Beat frequency F1 = 10.0 Hz

Find:

Speed of the approaching train

Computation:

Approaching frequency F2 = 220 + 10.0 Hz

Approaching frequency F2 = 230 Hz

Doppler shift;

F = [(v+v0)/(v-vS)]F0

230 = [(340+v0)/(340-0)]220

V0 = 15.45 m/s

Speed of the approaching train = 15.45 m/s

Answer:

28.3 m/s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 30°

Maximum height (H) = 10 m

Acceleration due to gravity (g) = 10 m/s²

Initial velocity (u) =?

Thus, we can obtain the minimum velocity cannon ball by using the following formula:

H = u²Sine² θ / 2g

10 = u² × (Sine 30)² / 2× 10

10 = u² × (0.5)² / 20

10 = u² × 0.25 / 20

10 = u² × 0.0125

Divide both side by 0.0125

u² = 10/ 0.0125

u² = 800

Take the square root of both side

u = √800

u = 28.3 m/s

Therefore, the minimum speed of the cannon ball is 28.3 m/s

Answer:

the lowest frequency f of the sound wave is 214.375 Hz

Explanation:

The computation of the lowest frequency f of the sound wave is shown below;

Length = L= 80 cm

= 0.8 m

V = 343 m/s (sound speed in air )

Now

V1 = n V ÷ 2 L

= 1 × 343 ÷ 2 × 0.8

V1 = 214.375 Hz

Hence, the lowest frequency f of the sound wave is 214.375 Hz

We simply applied the above formula so that the correct value could come

And, the same is to be considered  

Answer: the external agent must do work equal to -1.3 × 10⁻⁸ J

Explanation:

Given that;

Mass M1 = 7.0 kg

r = 3.0/2 m = 1.5 m

Mass M2 = 21 kg

we know that G = 6.67 × 10⁻¹¹ N.m²/kg²

work done by an external agent W = -2GM2M1 / r

so we substitute

W = (-2 × 6.67 × 10⁻¹¹ × 21 × 7) / 1.5

W = -1.96098 × 10⁻⁸ / 1.5

W = -1.3 × 10⁻⁸ J

Therefore the external agent must do work equal to -1.3 × 10⁻⁸ J

Answer:

T = 692.42 N

Explanation:

Given that,

Mass of hammer, m = 8.71 kg

Length of the chain to which an athlete whirls the hammer, r = 1.5 m

The angular sped of the hammer, [tex]\omega=1.16\ rev/s=7.28\ rad/s[/tex]

We need to find the tension in the chain. The tension acting in the chain is balanced by the required centripetal force. It is given by the formula as follows :

[tex]F=m\omega^2r\\\\=8.71\times (7.28)^2\times 1.5\\\\=692.42\ N[/tex]

So, the tension in the chain is 692.42 N.

Complete question:

Please find the image uploaded for the diagram.

Answer:

The magnitude of the current is 6 A, in the direction of Q to P.

Explanation:

Given;

length of the wire, l = 0.25 m

magnetic field strength, B = 40 mT = 0.04 T

Magnitude of the magnetic force, F = 60 mN = 0.06 N

The magnitude of the current in the conductor is given as;

F = BIL

Where;

I is the current induced in the conductor

I = F / BL

I = (0.06) / (0.04 x 0.25)

I = 6 A

Th current will move in direction of Q to P

Answer:

no

Explanation:

Answer:

Seventy-Five Metres

Explanation:

You can solve this simply by multiplying 15 m/s by 5 seconds

[tex]15 \frac{m}{s} * 5s\\=75\frac{ms}{s}\\= 75m[/tex]

The correct answer is a = 539.68 m/[tex]s^2[/tex]

A push or pull that one thing applies to another is known as force. An object is considered to exert a force when it accelerates through space because acceleration is the rate at which an item's speed changes. The second law of motion of Newton explains this concept.

The equation F=ma, where "F" stands for force, "m" for mass, and "a" for acceleration, illustrates the link between force and acceleration.

Force applied = F = 3.40 N

The bullet's weight = m = 6.30 g = 0.0063 kg

The bullet's rate of acceleration can be expressed as = a = F/m

a = 3.40 / 0.0063

a = 539.68 m/[tex]s^2[/tex]

To learn more about acceleration refer the link:

https://brainly.com/question/2303856

#SPJ9

Answer:

0.000314 Am²

6.049*10^-7 T

Explanation:

A

From the definitions of magnetic dipole moment, we can establish that

= , where

= the magnetic dipole moment in itself

= Current, 100 A

= Area, πr² (r = diameter divided by 2). Converting to m², we have 0.000001 m²

On solving, we have

= ,

= 100 * 3.14 * 0.000001

= 0.000314 Am²

B

= (0)/4 * 2 / ³, where

(0) = constant of permeability = 1.256*10^-6

z = 4.7 cm = 0.047 m

B = 1.256*10^-6 / 4*3.142 * [2 * 0.000314/0.047³]

B = 1*10^-7 * 0.000628/1.038*10^-4

B = 1*10^-7 * 6.049

B = 6.049*10^-7 T

Answer: D

Explanation:

Answer:

Explanation:

The Net Force

The net force is defined as the sum of all the forces acting on a body at a certain moment.

Recall some basic concepts and equations to solve the problem.

If no external forces are applied in the vertical direction, the weight of the object and the normal force have the same magnitude and point to opposite directions.

The friction force is defined as:

[tex]Fr_k=\mu_k N[/tex]

[tex]Fr_s=\mu_s N[/tex]

Where the subindices k and s are referred as to the kinetic and static friction forces respectively.

The condition for the object to move is that the applied force is greater than the friction force.

The crate of oranges has a mass of 30 Kg, thus its weight is:

W = m.g = 30 * 9.8 = 294 N

The normal force is:

N = W = 294 N

The kinetic friction is calculated as:

[tex]Fr_k=0.52* 294[/tex]

[tex]Fr_k=152.88\ N[/tex]

The static friction is calculated as:

[tex]Fr_s=0.62* 294[/tex]

[tex]Fr_s=182.28\ N[/tex]

The applied force is 200 N and it's greater than both the kinetic and the static friction forces, thus the crate is definitely moving at a certain positive acceleration.

Answer:

(b) 2.40 x [tex]10^{3}[/tex] kg/s

Explanation:

Given that: Total mass of the rocket = 3.003 x [tex]10^{5}[/tex] kg

acceleration of the rocket = 36.0 m[tex]s^{-2}[/tex]

speed of the exhausted gases = 4.503 x [tex]10^{3}[/tex] m/s

Rate at which rocket was initially burning fuel = [tex]\frac{mass}{time}[/tex]

But,

time = [tex]\frac{velocity}{acceleration}[/tex]

       = [tex]\frac{4.503*10^{3} }{36.0}[/tex]

       = 125.0833 s

So that;

Rate at which rocket was initially burning fuel = [tex]\frac{3.003*10^{5} }{125.0833}[/tex]

        = 2400.8001

        = 2.40 x [tex]10^{3}[/tex] kg/s

Therefore, the initial rate at which the rocket burn fuel is 2.40 x [tex]10^{3}[/tex] kg/s.