2. A 15.0 N force is used to hold a mass on a string at an angle of 32° from the vertical, as in the
below diagram.
32
F = 15.0 N
What is the mass? (ans: 2.9 kg)
As u can see I have the answer so plz just show ALL STEPS

Answers

Answer 1

Answer: 2.9kg

let mass be=m ; F=15N; Sin(theta)=32°;g =9.8kgm/s^2.

represent the mass. By resolving forces, we have:

F=mgSin(theta)

15N=mgSin(32°)

m=15N/g×Sin(32°)

m=15N/9.8m/s^2÷Sin(32°)

=2.888kg2.888

≈2.9kg

Answer 2

Mass on a string will be 2.9 kg

What is equilibrium condition?

Object at equilibrium is the condition in which all the forces are balanced and the object will not accelerate and net force on the object will be zero

given

mass = ?

Force = F = 15N

angle from the vertical = 32°

according to newton' s law of equation, because of equilibrium condition

mg sin(32°) - F =0

m(9.8) (0.5299) - 15 =0

5.19302 m = 15

m= 15/5.19302 = 2.8885 = 2.9 kg

Mass on a string will be 2.9 kg

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Related Questions

Suppose a 60-turn coil lies in the plane of the page in a uniform magnetic field that is directed out of the page. The coil originally has an area of 0.325 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.60 T

Answers

Answer:

 emf = 312 V

Explanation:

In this exercise the electromotive force is asked, for which we must use Faraday's law

           emf =  [tex]- N \frac{d \Phi }{dt}[/tex]- N dfi / dt

           Ф = B. A = B A cos θ

bold type indicates vectors.

They indicate that the magnetic field is constant, the angle between the normal to the area and the magnetic field is parallel by local cosine values ​​1

It also indicates that the area is reduced from  a₀ = 0.325 me² to a_f = 0 in a time interval of ΔT = 0.100 s, suppose that this reduction is linear

            emf = -N B [tex]\frac{dA}{dT}[/tex]

            emf = - N B (A_f - A₀) / Dt

we calculate

           emf = - 60 1.60 (0 - 0.325) /0.100

           emf = 312 V

The direction of this voltage is exiting the page

An astronaut on the moon drops a rock from rest. The rock falls 0.8m in one second of falling time. If the dropped rock fell for a total of two seconds of time instead of 1 second, then the distance traveled would be:____.
A) The same.
B) Doubled.
C) Tripled.
D) Quadruple.
E) None of the above.

Answers

Answer:

D) Quadruple.

Explanation:

We will use the second equation of motion to solve this problem:

[tex]s = v_it + \frac{1}{2}gt^2[/tex]

where,

s = distance travelled by the rock

vi = initial speed of rock = 0 m/s

t = time taken

g = acceleration due to gravity on the surface of the moon

Therefore,

[tex]s = (0\ m/s)t+\frac{1}{2}gt^2\\\\s =\frac{1}{2}gt^2[/tex]----------- equation (1)

Now, we double the time:

[tex]s' = \frac{1}{2}g(2t)^2\\\\s' = 4(\frac{1}{2}gt^2)[/tex]

using equation (1):

s' = 4s

Hence, the correct option is:

D) Quadruple.

I’m a photoelectric effect, which property of the incident light determines how much kinetic energy the ejected electrons have ?
A) brightness
B) frequency
C) size of the beam
D) none of the above

Answers

Answer:

b = frequency

A 500 kg rocket sled is coasting in reverse at 10 m/s (to the left). It then turns on its rocket engines for 10.0 s, with a thrust of 1500 N (to the right). What is its final velocity? (Remember velocity has magnitude and direction)

Answers

Explanation:

F = ma

[tex]a = \frac{f}{m} [/tex]

[tex]a = \frac{1500}{500} = 3[/tex]

[tex]a = \frac{v2 - v1}{t} [/tex]

[tex]3 = \frac{v2 - 10}{10} [/tex]

v2 (final) = 40 m/s to the right direction

A friend lends you the eyepiece of his microscope to use on your own microscope. He claims that since his eyepiece has the same diameter as yours but twice the focal length, the resolving power of your microscope will be doubled. Is his claim valid? Explain.

Answers

Answer:

The resolving power remains same.

Explanation:

The resolving power of the lens is directly proportional to the diameter of the lens not on the focal length.

As the diameter is same but the focal length is doubled so the resolving power remains same.

Explain how blood circulation takes place in humans?​

Answers

Blood comes into the right atrium from the body, moves into the right ventricle and is pushed into the pulmonary arteries in the lungs. After picking up oxygen, the blood travels back to the heart through the pulmonary veins into the left atrium, to the left ventricle and out to the body's tissues through the aorta.

Hope it helps you

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A positron is a particle that has the same mass but opposite charge of an electron. An electron and a positron are shot directly toward each other by a particle accelerator. They start very far from each other, each with a kinetic energy of

Answers

Complete Question

The Complete Question is Attached below

Answer:

[tex]E_t=2.14*10^{-13}J[/tex]

Explanation:

From the question we are told that:

Kinetic energy [tex]K.E=5*10^{-14}[/tex]

Generally, the equation for Total Energy is mathematically given by

[tex]E_t=K.E+2me[/tex]

Where

Mass of an electron in MeV

[tex]m=0.510 998 950 00 MeV[/tex]

And

Electron Charge

[tex]1.6 * 10^{-19}[/tex]

Therefore

[tex]E_t=K.E+2me[/tex]

[tex]E_t=5*10^{-14}*0.510 998 950 00*1.6 * 10^{-19}[/tex]

[tex]E_t=2.14*10^{-13}J[/tex]

A capacitor consists of two parallel conducting plates, each of area 0.4 m2 and separated by a distance of 2.0 cm. Assume there is air between the plates. While connected to a battery the electric field within the plates is 500 N/C. The potential difference between the plates is: ________
a) 5.0 V
b) 10 V
c) 30 V
d) 20 V

Answers

Answer:

check photo

Explanation:

The cation that is reabsorbed from the urine in response to aldosterone

Answers

Answer:

If decreased blood pressure is detected, the adrenal gland is stimulated by these stretch receptors to release aldosterone, which increases sodium reabsorption from the urine, sweat, and the gut. This causes increased osmolarity in the extracellular fluid, which will eventually return blood pressure toward normal.

5. Steve is driving in his car to take care of some errands. The first errand has him driving to a location 2 km East and 6 km North of his starting location. Once he completes that errand, he drives to the second one which is 4 km East and 2 km South of the first errand. What is the magnitude of the vector that describes how far the car has traveled from its starting point, rounded to the nearest km?​

Answers

Answer:

gshshs

Explanation:

hshsksksksbsbbshd

list at least types of motion​

Answers

Answer:

These four are rotary, oscillating, linear and reciprocating. Each one moves in a slightly different way and each type of achieved using different mechanical means that help us understand linear motion and motion control.

(I got this off the web so credits to the rightful owner and I hope you have good day :)

A 1.5kg block slides along a frictionless surface at 1.3m/s . A second block, sliding at a faster 4.3m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.0m/s . What was the mass of the second block?

Answers

Answer:

The mass of the second block=0.457 kg

Explanation:

We are given that

m1=1.5 kg

v1=1.3m/s

v2=4.3 m/s

V=2.0 m/s

We have to find the mass of the second block.

[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]

Let m2=m

Substitute the values

[tex]1.5(1.3)+m(4.3)=(1.5+m)(2)[/tex]

[tex]1.95+4.3m=3+2m[/tex]

[tex]4.3m-2m=3-1.95[/tex]

[tex]2.3m=1.05[/tex]

[tex]m=\frac{1.05}{2.3}[/tex]

[tex]m=0.457 kg[/tex]

Hence,  the mass of the second block=0.457 kg

calculate the rate of loss of heat through a glass window of area 200 CM square and thickness 0.5 CM where temperature inside is 35 degree Celsius and outside is -5 degree Celsius conductivity of Glass is 2.2 into 10 to the power 3 cal per s per cm per k .

Answers

Answer:

The inner and outer surfaces of a 0.5-cm thick 2-m by 2-m window glass in winter are 10°C and 3°C, respectively. If the thermal conductivity of the glass

Explanation:

A force of 16.88 N is applied tangentially to a wheel of radius 0.340 m and gives rise to an angular acceleration of 1.20rad / (s ^ 2) . Calculate the rotational inertia of the wheel. A. 2.77 kg - m ^ 2 B. 0.73 kg - m ^ 2 C. 4.41 kg - m ^ 2 O. 4.78 kg - m ^ 2

Answers

Given.

force = 16.88 N is

radius = 0.340m

an angular acceleration = 1.20rad/s^2

the formula for torque is

F*r = I*a

where I is moment of inertia

16.88*.34 = I*1.2

I = 4.78Kg-m^2

so rotational inertia I = 4.78Kg-m^2

what is liquid pressure and its si unit?

Answers

The SI unit of pressure is the pascal: 1Pa=1N/m2 1 Pa = 1 N/m 2 . Pressure due to the weight of a liquid of constant density is given by p=ρgh p = ρ g h , where p is the pressure, h is the depth of the liquid, ρ is the density of the liquid, and g is the acceleration due to gravity.

Why is it advised not to hold the thermometer by its bulb while reading it?

Answers

Because your body heat might change the temperature

if you jog at a speed of 1.5m/s for 20 seconds how far di you travel

Answers

Answer: 30m

Explanation:

Given:

Speed: 1.5m/s

Time: 20 seconds

Distance = speed × time

Distance = 1.5 × 20

= 30m

Therefore you will travel 30m

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What is the frequency of a wavelength

Answers

the longer the wavelength, lower the frequency. In the same manner, shorter the wavelength, higher will be the frequency.

Wavelength is inversely proportional to frequency.

In an equation, wavelength is represented by the Greek letter lambda (λ). Depending on the type of wave, wavelength can be measured in meters, centimeters, or nanometers (1 m = 109 nm). The frequency, represented by the Greek letter nu (ν), is the number of waves that pass a certain point in a specified amount of time.

Hope this helps :))

physics approach to study macromoelcues at nanoscales
in detail plx

Answers

Answer:

Abstracto

Los ácidos nucleicos y las proteínas comprenden una red de biomacromoléculas que almacenan y transmiten información que sustenta la vida de la célula. El estudio de estos mecanismos es un campo llamado biología molecular. El desarrollo de esta ciencia siempre ha ido acompañado de avances técnicos que permiten romper barreras metodológicas para probar hipótesis novedosas. Entre los métodos disponibles para los biólogos moleculares, destacan cinco: electroforesis, secuenciación, clonación, transferencia y reacción en cadena de la polimerasa. Su impacto llega a la genética, la medicina y la biotecnología. Aquí, se revisan la relevancia histórica, los fundamentos técnicos y las tendencias actuales de estos cinco métodos esenciales. La revisión pretende ser útil tanto para estudiantes como para científicos profesionales que buscan adquirir conocimientos avanzados sobre el valor de estos métodos para investigar los mecanismos moleculares que sostienen la vida.

. A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?

Answers

Answer:

4

Explanation:

Which factor affects kinetic energy but not potential energy?

Answers

Answer:

mass

Explanation:

because as the mass increase kinetic energy also increase

Transistor circuits are sometimes referred to as switching circuits - Why is this?​

Answers

Explanation:

One of the most common uses for transistors in an electronic circuit is as simple switches. In short, a transistor conducts current across the collector-emitter path only when a voltage is applied to the base. When no base voltage is present, the switch is off. When base voltage is present, the switch is on.

A simple pendulum consists of a ball of mass 3 kg hanging from a uniform string of mass 0.05 kg and length L. If the period of oscillation of the pendulum is 2 s, determine the speed of a transverse wave in the string when the pendulum hangs vertically.

Answers

Answer:

v = 3.12 m/s

Explanation:

First, we will find the length of the string by using the formula of the time period:

[tex]T = 2\pi \sqrt{\frac{l}{g}}\\\\l = \frac{T^2g}{4\pi^2}\\\\[/tex]

where,

l = length of string = ?

T = time period = 2 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]l = \frac{(2\ s)^2(9.81\ m/s^2)}{4\pi^2}\\\\l = 0.99\ m[/tex]

Now, we will find tension in the string in the vertical position through the weight of the ball:

T = W = mg = (3 kg)(9.81 m/s²)

T = 29.43 N

Now, the speed of the transverse wave is given as follows:

[tex]v=\sqrt{\frac{Tl}{m}}\\\\v=\sqrt{\frac{(29.43\ N)(0.99\ m)}{3\ kg}}\\\\[/tex]

v = 3.12 m/s

A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if the slit width is doubled?
a. It will become four times as large.
b. It will double.
c. It will be cut in half.
d. It will become eight times as large.
e. It will be cut to one-quarter its original size.

Answers

Answer:

c : it wil be cut in half.

The pattern is formed on a distant screen so we can use the Fraunhofer difracction for a single slit. The formula of the width of the central bright spot is given by Δx = (2λz)/a, where λ is the wavelength and a is the width of the slit. So if the inicial width (a_1) is doubled (a_2= 2 x a_1),the width of the central spot will be cut in half Δx = (2λz)/a_2 = (2λz)/2xa_1 .

Rachel has good distant vision but has a touch of presbyopia. Her near point is 0.60 m. Part A When she wears 2.0 D reading glasses, what is her near point

Answers

Answer:

The right answer is "0.273 m".

Explanation:

Given:

Power (P),

[tex]\frac{1}{f} = 2D[/tex]

Near point,

u = 0.6 m

As we know,

⇒ [tex]\frac{1}{v} -\frac{1}{u}=\frac{1}{f} = 2[/tex]

By substituting the values, we get

⇒ [tex]\frac{1}{v} -\frac{1}{0.6} =2[/tex]

            [tex]\frac{1}{v}=2+\frac{1}{0.6}[/tex]

            [tex]\frac{1}{v} =\frac{1.2+1}{0.6}[/tex]

            [tex]\frac{1}{v}=\frac{2.2}{0.6}[/tex]    

By applying cross-multiplication, we get

          [tex]0.6=2.2 \ v[/tex]

            [tex]v = \frac{0.6}{2.2}[/tex]

      [tex]S_{near} = 0.273 \ m[/tex]

b. Projectile on cliff (range)
An object of mass 5 kg is projected at an angle of 25° to the horizontal with a speed of 22 ms-1 from the top of the cliff.
The height of the cliff is 21 m. Take g, the acceleration due to gravity, to be 9.81 ms2
How far horizontally (to 1 decimal place) from the base of the cliff does the object land?

Answers

Answer:

x = 41.28 m

Explanation:

This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.

Let's start by using trigonometry to find the initial velocity

         cos 25 = v₀ₓ / v₀

         sin 25 = Iv_{oy} / v₀

         v₀ₓ = v₀ cos 25

         v_{oy} = v₀ sin 25

         v₀ₓ = 22 cos 25 = 19.94 m / s

         v_{oy} = 22 sin 25 = 0.0192 m / s

let's use movement on the vertical axis

         y = y₀ + v_{oy} t - ½ g t²

     

when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m

         0 = 21 + 0.0192 t - ½ 9.81 t²

         4.905 t² - 0.0192 t - 21 = 0

         t² - 0.003914 t - 4.2813 =0

we solve the quadratic equation

        t = [tex]\frac{ 0.003914\ \pm \sqrt{0.003914^2 + 4 \ 4.2813 } }{2}[/tex]

        t = [tex]\frac{0.003914 \ \pm 4.13828}{2}[/tex]

        t₁ = 2.07 s

        t₂ = -2.067 s

since time must be a positive scalar quantity, the correct result is

        t = 2.07 s

now we can look up the distance traveled

         x = v₀ₓ t

         x = 19.94  2.07

         x = 41.28 m

Please help wil give brainiest & 40p.

Answers

Answer:

This is the answer

Explanation:

You can find the ans in the photo I attached.

The pressure of sea water increases by 1.0atm for each 10m increase in the depth, by what percentage is the density of water increased in the deepest ocean of water of 12km. Compressibility is 5.0×10^-5 atm

Answers

The percentage by which the water density increased is 4.1[tex]\mathbf{\overline 6}[/tex] %

The known values are;

The increase in pressure per 10 meter increase in depth = 1.0 atm

The depth of the deepest ocean = 12 km = 12,000 m

The compressibility of the ocean = 5.0 × 10⁻⁵ 1/atm

The unknown

The percentage the density of water increased in the deepest ocean

Strategy;

Find the pressure at the deepest point of the deepest ocean and apply the compressibility

We have;

[tex]\mathbf{Compressibility = \dfrac{1}{V} \times \dfrac{\partial V}{\partial p}}[/tex]

The change in pressure, [tex]\partial p[/tex] = (12,000 m/(10 m)) × 1.0 atm = 1,200 atm

Therefore, we have for one cubic meter of water

[tex]\mathbf{5.0 \times 10^{-5} \ atm^{-1} = \dfrac{1}{1 \, m^3} \times \dfrac{\partial V}{1,200 \, atm}}[/tex]

Therefore;

[tex]\mathbf{\partial}[/tex]V = 5.0 × 10⁻⁵ atm⁻¹ × 1 m³ × 1,200 atm = 0.06 m³

The new volume = V - [tex]\mathbf{\partial}[/tex]V

∴ The new volume = 1 m³ - 0.06 m³ = 0.94 m³

The initial density = mass/(1 m³)

The new density = mass/(0.96 m³)

The percentage increase in density, [tex]\partial[/tex]ρ%, is given as follows;

[tex]\mathbf{\partial p \% = \dfrac{ \dfrac{Mass}{0.96 \ m^3} - \dfrac{Mass}{1 \ m^3} }{ \dfrac{Mass}{1 \ m^3}} \times 100 = \dfrac{25}{6} \% = 4.1 \overline 6 \%}[/tex]

∴  [tex]\mathbf{\partial}[/tex]ρ% =  4.1[tex]\mathbf {\overline 6}[/tex] %

The percentage by which the water density increased, [tex]\partial[/tex]ρ% = 4.1[tex]\mathbf{\overline 6}[/tex] %

Learn more about compressibility here;

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why the stone moves away when the string is broken rotation​

Answers

Answer:

When a stone is going around a circular path, the instantaneous velocity of stone is acting as tangent to the circle. When the string breaks, the centripetal force stops to act. Due to inertia, the stone continues to move along the tangent to circular path. So, the stone flies off tangentially to the circular path

Answer:

when the string's rotation is broken, there will be no centripetal force to keep the stone stationary. Thus, the stone will flung away when the rotation is stopped

The coefficients of friction between a race cars tyres and the track surface are

Answers

the question is about tyres of a race car, which are made of rubber and will be in contact with a race track, which is generally made from asphalt, the static coefficient of friction is in the range of (0.5–0.8), in dry conditions (Source: Friction and Friction Coefficients ).

Explanation:

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