Net force is the sum of all the forces acting on an object. If a spring balance pulls on a body with a force of 10 N, and friction acts on the body in the opposite direction with a force of 1 N, the net force would be 9 N in the direction of the spring balance (10 N – 1 N = 9 N).What is the net force acting on the object when the spring balance pulls the rope with a force of 25 N and friction acts on the body with a force of 20N?

Answers

Answer 1

Answer:

Answer:

Explanation:

(25 N - 20 N = 5 N)

Related Questions

Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4 = 407 N , and MA = 1504 N⋅m . Express the Cartesian components of the resultant force and the couple moment in newtons and newton-meters to three significant figures separated by commas.
A 50.0-kg box is being pulled along a horizontal surface by means of a rope that exerts a force of 250 n at an angle of 32.0° above the horizontal. the coefficient of kinetic friction between the box and the surface is 0.350. what is the acceleration of the box?
Two charged particles are located on the x axis. The first is a charge +Q at x = −a. The second is an unknown charge located at x = +3a. The net electric field these charges produce at the origin has a magnitude of 2keQ/a2 . Explain how many values are possible for the unknown charge and find the possible values.
Consider position [x] = L, time [t] = T, velocity [v] = L/T and acceleration [a] = L/T 2 . Find the exponent A in the equation v = a^2 t^ A /x
The knot at the junction is in equilibrium under the influence of four forces acting on it. The F force acts from above on the left at an angle of α with the horizontal. The 5.7 N force acts from above on the right at an angle of 50◦ with the horizontal. The 6.2 N force acts from below on the right at an angle of 44◦ with the horizontal. The 6.7 N force acts from below on the left at an angle of 43◦ with the horizontal.1. What is the magnitude of the force F?2. What is the angle a of the force F in the figure above?

A machine can make doing work easier by reducing the force exerted, changing the distance over which the force is exerted, or changing the direction of the force.True OR False

HELP ME!!!!!!¡!!!!

Answers

I believe the correct answer is true. A machine can make doing work easier by reducing the force exerted, changing the distance over which the force is exerted, or changing the direction of the force. Hope this answer the question.

An element emits light at two nearly equal wavelengths, 577 nm and 579 nm If the light is normally incident on a diffraction grating with 2000 lines/cm., what is the distance between the 3rd order fringes of the two wavelengths on a screen 1 m from the grating?

Answers

Answer:

Explanation:

d = width of slit = 1 / 2000 cm =5 x 10⁻⁶ m

Distance of screen D = 1 m.

wave length λ₁ and λ₂ are 577 x 10⁻⁹ and 579 x 10⁻⁹ m.respectively.

distance of third order bright fringe = 3.5 λ D/d

for 577 nm , this distance = 3.5 x 577 x 10⁻⁹ x 1 /5 x 10⁻⁶

= .403 m = 40.3 cm

For 579 nm , this distance = 3.5 x 579 x 1 / 5 x 10⁻⁶

= 40.5 cm

Distance between these two = 0.2 cm.

Every few years, winds in Boulder, Colorado, attain sustained speeds of 45.0 m/s (about 100 mi/h) when the jet stream descends during early spring. show answer No Attempt Approximately what is the force due to the Bernoulli effect on a roof having an area of 205 m2? Typical air density in Boulder is 1.14 kg/m3 , and the corresponding atmospheric pressure is 8.89 × 104 N/m2 . (Bernoulli’s principle assumes a laminar flow. Using the principle here produces only an approximate result, because there is significant turbulence.)

Answers

Answer:

The force exerted on the roof is $F =2.37*10^(5)N$

Explanation:

From the question we are told that

The speed of the wind is $v = 45.0 m/s$

The area of the roof is $A = 205 m^2$

The air density of Boulder is $\rho = 1.14 kg / m^3$

The atmospheric pressure is $P_(atm) = 8.89 * 10^(4) N/ m^2$

For a laminar flow the Bernoulli’s principle is mathematically represented as

$P_1 + (1)/(2) \rho v_a ^2 + \rho g h_a = P_2 + (1)/(2) \rho v_b ^2 + \rho h_b$

Where $v_1$ is the speed of air in the building

$v_b$ is the speed of air outside the building

$P_1 \ and \ P_2$ are the pressure of inside and outside the house

$h_a \ and \ h_b$ are the height above and below the roof

Now for $h_a = h_b$

The above equation becomes

$P_1 + (1)/(2) \rho v_a ^2 = P_2 + (1)/(2) \rho v_b ^2$

$P_1 - P_2 = (1)/(2) \rho (v_b^2 - v_a^2)$

Since pressure is mathematically represented as

$P = (F)/(A )$

The above equation can be written as

$F = (1)/(2) \rho ( v_b^2 - v_a ^2 ) A$

The initial velocity is 0

Substituting value

$F = (1)/(2) (1.14) [(45^2 - 0^2 ) ](205)$

$F =2.37*10^(5)N$

A wire with mass 90.0 g is stretched so that its ends are tied down at points 98.0 cm apart. The wire vibrates in its fundamental mode with frequency 60.0 Hz and with an amplitude of 0.300 cm at the antinodes. Part A What is the speed of propagation of transverse waves in the wire

Answers

Answer:

118 m/s

Explanation:

Given :

We know that

$f\ =\ (1)/(2l) \sqrt{(T)/(U) }$ ......Eq(1)

Where $\sqrt{(T)/(u) }$ =v

l=length

f=frequency

l= 98.0 cm= 0.98 m

f=60.0 Hz

Now from the Eq(1)

$f\ =\ (v)/(2l)$

This equation can be written as

v=2fl.............Eq(2)

Putting the value f and l in Eq(2)

v=2*60*0.98

v=117.6 m/s ~ 118 m/s

A disk of mass 5 kg and radius 1m is rotating about its center. A lump of clay of mass 3kg is dropped onto the disk at a radius of 0.5m , sticking to the disk. If the system is rotating with an angular velocity of 11 rad/s, what is the final angular momentum of the disk h the clay lump?wit? ( Idisk = MR^2/2)

Answers

Answer:

$27.5 kgm^2/s$

Explanation:

We can solve for the final angular velocity of the system using the law of momentum conservation

$I_1\omega_1 = I_2\omega_2 = M_2$

Where $I_1 = MR^2/2 = 5*1^2/2 = 2.5 kgm^2$ is the moments of inertia of the disk before. $I_2 = I_1 + mr^2 = 2.5 + 3*0.5^2 = 2.5 + 0.75 = 3.25 kgm^2$ is the moments of inertia of the disk after (if we treat the clay as a point particle). $\omega_1 = 11rad/s$ is the angular speed before.

$2.5*11 = M_2$

$M_2 = 27.5 kgm^2/s$

So the final momentum of the system is 27.5 kgm2/s

Answer:

The final angular momentum is 35.75 kg.m²/s

Explanation:

Given;

mass of disk, M = 5 kg

radius of disk, R = 1 m

mass of clay, M = 3 kg

radius of clay, R = 0.5 m

final angular momentum, $\omega _f$ = 11 rad/s

Final angular momentum angular momentum of the disk that the clay lumped with;

$P = I_f\omega_f$

where;

$I_f$ is the final moment of inertia

$I_f = I_(disk) + I _(sand)\n\nI_f = (M_DR^2)/(2) + M_SR^2\n\nI_f = (5*1^2)/(2)+ 3*0.5^2\n\nI_f = 2.5 + 0.75=3.25 \ kg.m^2$

Final angular momentum of the disk;

= $I_f \omega_f$

= 3.25 x 11 = 35.75 kg.m²/s

Therefore, the final angular momentum is 35.75 kg.m²/s

A block (0.50 kg) is attached to an ideal spring with a spring constant of 80 N/m, oscillating horizontally on a frictionless surface. The total mechanical energy is 0.12 J. (a) What is the greatest extension of the spring from its equilibrium length? (b) Now the block is traveling 2.00 m/s, and brought to rest by compressing a very long spring of spring constant 800.0 N/m. How much does the spring compress?

Answers

Answer:

a) x = 5.48 10⁻² m and b) 0.05 m

Explanation:

a) For a system in oscillatory motion the mechanical energy conserves and is described by the equation

Em = ½ k A²

Where k is the spring constant and at the amplitude of the movement

When the spring has the greatest extent, the kinetic energy is zero

Em = U = ½ k x²

Therefore, the amplitude of the movement is the same amplitude of the spring

Let's calculate

A = √ (2Em / k)

A = √ (2 0.12 / 80)

A = 0.0548 m = 5.48 10⁻² m

b) In this case the spring has kinetic energy that becomes elastic potential energy, let's calculate the mechanical energy before and after compressing the spring

Initial

Em = K = ½ m v²

Final

Em = Ke = ½ k x²

½ m v² = ½ k x²

x = √(m/k) v

x = 2 √(0.50 /800.0)

x = 0.05 m

Answer:

a) The greatest extension of the spring is 0.055 m

b) The spring compress 0.05 m

Explanation: