Pick the free body diagram that correctly matches the following situation. An object is not moving.

Answer:

Mag isip ka po wag Kang tanong ng tanong kac kami po Yung nahihirapan mag hanap ng answer

Answer:

Because it is the standard operational range for measuring the body temperature of a living individual.

Explanation:

When making a measuring equipment like a thermometer, you attempt to keep the reading range (Span) as small as possible so that the reading is as accurate as feasible. Because sensors are often rated in percent of span (in this example, span = 45 C–32 C = 13 degrees Celsius), One percent of 13 equals 0.13 degrees Celsius, thus if your sensor has a usual accuracy of 1%, your temperature reading is accurate to roughly 0.1 degrees Celsius.

Your precision would be 0.2 degrees Celsius if you doubled the span. The reading accuracy is now becoming unsatisfactory. Readability is another advantage of analog and mercury clinical thermometers. The markings for 0.1 deg C would be ten times closer together if a mercury thermometer had ten times the range (span). There is a limit to the device's readability.

A rifle recoils while firing a bullet. The speed of the rifle's recoil is small because the rifle has more mass than the bullet.

[tex]p_{1} =-p_2}[/tex]

[tex]m_{1} V_{1} =-M_{2} v_{2}[/tex]

Answer:

Explanation:

A rifle recoils while firing a bullet. The speed of the rifle's recoil is small  and typically fast because the rifle has more mass than the bullet.

Answer:

8 m/s

Explanation:

(16-0)/2 = 8 m/s

The magnitude of the vertical force that a concrete support exerts on the bridge at the far end is 201.25 N.

The force is an action of push or pull which makes a body to move or stop.

Given is the weight of tourist W = 805 N and L is the length of the bridge.

Let the reaction force at both the ends of the bridge is R1 and R2.

Taking the moments about point A,

R1 x 0 - W x L/4 +R2 x L =0

R2 = W/4

R2 = 805 /4

R2 = 201.25 N

Thus, the magnitude of the vertical force that a concrete support exerts on the bridge at the far end is 201.25 N.

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Given ,

Mass of the body (m) = 200g

Amplitude (A) = 20mm

Maximum force (F) = 0.6 N

Maximum velocity (vmax) =?

We know that

Fmax = mv2max/A

or , 0.6 = 200×10-3xv2max/20×10-3

or , 0.6×20/200= v2max

or , vmax = 0.24 m/s

∴ Maximum velocity = 0.24 m/s2

The value of maximum velocity will be 0.3464 m/s². Energy or work is equal to the product of force and displacement.

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

Mass of the body is,(m= 200g)

Amplitude is,(A  =20 mm)

Maximum force is,F= 0.6 N

To find;

Maximum velocity

Energy or work is equal to the product of force and displacement.

[tex]\rm KE = \frac{1}{2} mV_{max}^2 \\\\ \rm F_{max} \times A = \frac{1}{2} mV_{max}^2 \\\\ F_{max}= \frac{1}{2}\frac{ mV_{max}^2}{A} \\\\ 0.6= \frac{1}{2}\times \frac{ 0.2 \times V_{max}^2}{20 \times 10^{-3}} \\\\ V_{max}=0.3464 \ m/sec^2[/tex]

Hence,the value of maximum velocity will be 0.3464 m/s².

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Answer: reference point

Explanation: A reference point could be a put or question utilized for comparison to decide in case something is in movement.

Answer:

529.2 joules

Explanation:

Work done = force x distance = mass x gravity x distance

= 36x9.8 x1.5 =529.2 joules

Answer:

I think the letter B. But I'm not sure pero I try mo Lang po baka Yan po ang tamang sagot kac Yan po Yung nilagay na sagot saakin e Kaya try mo Lang po

8.3×10 to the power minus 7

Explanation:

lambda equals velocity over frequency.

GPE = mgh

GPE = 57 x 9.8 x 381 = 2.12 x 10⁵ J

The change in momentum of the object during the half period is -10 kgm/s.

The change in momentum of the object during the one period is 0.

The change in momentum of the object during the half period is calculated as follows;

ΔP(¹/₂) = Pf - Pi

where;

ΔP(¹/₂) = (-5 x ) - (5 x 1)  

ΔP(¹/₂) = -10 kgm/s

ΔP(1) = Pf - Pi

ΔP(1) = (5 x 1) - (5 x 1) = 0

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Answer:

wait one second im about to go get some milk im thirsty

Explanation:

do you want any chicken on your salad

Answer:

Concave lens

.............

[tex]\Large{\green}\fcolorbox{blue}{cyan}{\bf{\underline{\red{\color{red}Answer}}}}[/tex]

2300 N upward

Explanation:

The force by weight of balloon = 1000 N

The atmospheric force experienced by balloon = 3300 N

While weight of balloon is attracted by gravity so the force of weight will act downward

And the atmospheric force gives upward lifting force to the balloon

Net force = atmospheric force - weight of balloon

= 3300 - 1000

= 2300 N

As the atmospheric force is more dominating the balloon will goes upward with 2300 N force

Answer:

0.315C

Explanation:

Force/Electric field

7520N/23900C

Answer:

290.4 joules

Explanation:

As the formula to find kinetic energy is 1/2 mv^2.

So if we put the given data in it,the answer comes 290.4 joules

#Jennifer

Answer:

The specific heat capacity is defined as the quantity of heat (J) absorbed per unit mass (kg) of the material when its temperature increases 1 K (or 1 °C), and its units are J/(kg K) or J/(kg °C).

Answer:

M g - T =  M a     acceleration of hanging block   (M = 2)

2 g - T = 2 a        where M = 2

T = 2 g - 2 a

T = 8 * a + 12 N       acceleration of blocks on table

2 g - 2 a = 8 a + 12

10 a = 2 g - 12  

a = .8 m/s^2   (assuming g = 10 m/s^2)

T = 2 g - 2 a = 2 * 10 - 2 * .8 = 20 - 1.6 = 18.4 N

Check: For blocks on table

a = (18.4 - 12) / 8 = 6.4 / 8 = .8 m/s^2

Answer:

See below

Explanation:

The units of the specific heat hint at how to solve these types of problems

.88 J/(g-C)   * 10 g  *  (100- 50 C) = 440 J  

                                  ( see how the g and the C 'cancel out' ?)

Answer:

true

Explanation:

Answer:

0.5

Explanation:

I think it is this beacause if u have the answer you have that

The concept of the centrifugal force is used for producing an effective imitation of gravity. It is an important concept in the mechanics.

when any body is executing circular motion about a fixed axis, an outward force act on that object to balance the body and to execute the circular motion. The outward force is known as the centrifugal force.

It is the pseudo force act in order to balance the centripetal force acting inward during the circular motion.

The centrifugal force is comes into the picture when any body executes circular motion about a fixed axis.

To simulate gravity for the inhabitants, the concept of the centrifugal force is used.

Hence, the concept of the centrifugal force is used for producing an effective imitation of gravity.

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1)The period of these ripples will be 0.2 sec.

2)The time period of sound wave will be 100 Hz.

3)The frequency of the sound wave will be 5×10⁶ Hz.

Frequency is defined as the number of repetitions of a wave occurring waves in 1 second.

Frequency is given by the formula as,

Frequency of  ripples (f₁)=5 Hz

The period of these ripples will be;

T₁=1/f₁

T₁=1/5

T₁=0.2 sec

The time period of sound wave (T)=0.01 seconds

The time period of sound wave;

f=1/T

f=1/0.01

f=100 Hz

Time period of A radio wave = 0.0000002 seconds

The frequency of the sound wave is;

f=1/T

f=1/0.0000002

f=5×10⁶ Hz

Hence,the period of these ripples will be 0.2 sec.

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Consider,

[tex]{:\implies \quad \displaystyle \sf \langle p\rangle =m\langle v(t)\rangle=m\int_{-\infty}^{\infty}x\bigg\{\dfrac{\partial \Psi^{*}(x,t)}{\partial t}\Psi (x,t)+\Psi^{*}(x,t)\dfrac{\partial \Psi (x,t)}{\partial t}\bigg\}dx}[/tex]

Multiply both sides by ih and simplification will yield

[tex]{:\implies \quad \displaystyle \sf ih\langle p\rangle =m\int_{-\infty}^{\infty}x\bigg[\Psi (x,t)\bigg\{\dfrac{h^2}{2m}\dfrac{\partial^{2}\Psi^{*}(x,t)}{\partial x^2}-V(x)\Psi^{*}(x,t)\bigg\}+\Psi^{*}(x,t)\bigg\{V(x)\Psi (x,t)-\dfrac{h^2}{2m}\dfrac{\partial^{2}\Psi (x,t)}{\partial x^2}\bigg\}\bigg]dx}[/tex]

Some simplification, Then Integrate by parts and then knowing the fact that the wave function vanishes for [tex]{\bf x\to \pm \infty}[/tex] will yield:

[tex]{:\implies \quad \displaystyle \sf \langle p\rangle =\dfrac{ih}{2}\int_{-\infty}^{\infty}\bigg\{\dfrac{\partial \Psi^{*}(x,t)}{\partial x}\Psi (x,t)-\Psi^{*}(x,t)\dfrac{\partial \Psi (x,t)}{\partial x}\bigg\}dx}[/tex]

Integrating by parts and knowing the same fact by some simplification will yield:

[tex]{:\implies \quad \displaystyle \sf \langle p\rangle =-ih\int_{-\infty}^{\infty}\Psi^{*}(x,t)\dfrac{\partial \Psi (x,t)}{\partial x}dx}[/tex]

The momentum is thus contained within the wave function, so we can then deduce that:

[tex]{:\implies \quad \sf p\rightarrow -ih\dfrac{\partial}{\partial x}}[/tex]

[tex]{:\implies \quad \sf p^{n}\rightarrow \bigg(-ih\dfrac{\partial}{\partial x}\bigg)^{n}}[/tex]

[tex]{:\implies \therefore \quad \displaystyle \sf \langle p^{2}\rangle =-h^{2}\int_{-\infty}^{\infty}\Psi^{*}(x,t)\dfrac{\partial^{2}\Psi (x,t)}{\partial x^2}dx}[/tex]

Now the kinetic energy

[tex]{:\implies \quad \displaystyle \sf \langle K\rangle =\dfrac{\langle p^{2}\rangle}{2m}=\dfrac{-h^2}{2m}\int_{-\infty}^{\infty}\Psi^{*}(x,t)\dfrac{\partial^{2}\Psi (x,t)}{\partial x^2}dx}[/tex]

The classical formula for the total energy

[tex]{:\implies \quad \sf \dfrac{p^2}{2m}+V(x)=E}[/tex]

Multiplying this equation by [tex]{\sf \Psi (x,t)=\psi (x)exp\bigg(\dfrac{-iEt}{h}\bigg)}[/tex] and use the above equations and simplify it we will be having

[tex]{:\implies \quad \boxed{\bf{\dfrac{-h^2}{2m}\dfrac{d^{2}\psi (x)}{dx^{2}}+V(x)\psi (x)=E\psi (x)}}}[/tex]

This is the Famous Time-Independent Schrödinger wave equation

Note:- If I write all the explanation then the Answer box willn't allow me to submit the answer

Answer:

acts as a one-way switch for current

Answer:

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Explanation: