What is the force applied by the ground called? A. applied force B. friction C. gravity

Explanation:

F = ma

m = F/a

= 12N / 2m/s2

= 6kg

The mass of the object is 6kg.

A force is defined as an effect that can change the motion of an object by which an object with mass can change its velocity, that is, accelerate. Force can also be described simply as a push or pull. A force has both magnitude and direction which makes it a vector quantity.

According to the Newton's second law of motion is F = ma, or force is equal to mass times acceleration.  

Where, F= Force applied

m= mass of the object

a= acceleration

For above given information,

F= 12 N

Acceleration= [tex]2 m/s^2[/tex]

So, mass will be m=F/a

m= 12/2 = 6kg

Thus, the mass of the object is 6kg.

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Given :

A boy walked 22 m East then 43 m West , and then 3 meters East.

To Find :

What is the sum of the boys motion.

Solution :

We need to find the sum of all the motion of boy .

So , we need to find the distance covered by the boy .

We know , total distance covered is absolute sum of all the motion .

[tex]D=22+43+3\\\\D=68\ meters[/tex]

Therefore , total distance covered is 68 meters .

Hence , this is the required solution .

Answer:

It's 18m West.

Explanation:

Imagine you have a number line. your starting point was 22 east of origin. Then you went 43 meters west which leaves you at 21m west. Then you go back east 3 meters. Which leaves you at 18 m west. I hope this helps.

Answer:

v = 0.58 m/s

Explanation:

The velocity of an object is given as the ratio of the total distance traveled by the object to the time interval taken for traveling. Hence, we use the following formula to find the average velocity of the cyclist:

v = s/t

where,

v = average velocity = ?

s = distance traveled = final position - initial position =1200 m - 150 m =1050 m

t = time interval = (30 min)(60 s/1 min) = 1800 s

Therefore,

v = 1050 m/1800 s

v = 0.58 m/s

Answer:

average acceleration = 6 [tex]\frac{m}{s^2}[/tex]

Explanation:

Recall that the average acceleration [tex](a)[/tex]  is defined by the change in velocity from an initial velocity [tex](v_i)[/tex], to a final velocity [tex](v_f)[/tex] over the time (t) it took that change to happen. Then, in mathematical terms this is:

[tex]a=\frac{v_f-v_i}{t}[/tex]

with our information this becomes:

[tex]a=\frac{v_f-v_i}{t} = \frac{100-40}{10}=6\,\frac{m}{s^2}[/tex]

Answer:

The value is [tex]P_1 = 314645 \ Pa [/tex]

Explanation:

From the question we are told that

The height is [tex]h_2 = 10 m[/tex]

The height of the burning roof is [tex]k = 9 m[/tex]

The horizontal distance is [tex]d = 7 \ m[/tex]

The height of the truck is [tex]h_1 = 0.5 \ m[/tex]

Generally the time for the water to hit the roof from the hose is mathematically represented as

[tex]t = \sqrt{\frac{2 * (h_2 - k)}{g} }[/tex]

=> [tex]t = \sqrt{\frac{2 * (10 - 9)}{9.8} }[/tex]

=> [tex]t = 0.4518 \ s [/tex]

Generally the velocity of the water is mathematically evaluated as

[tex]v_2 = \frac{d}{t}[/tex]

[tex]v_2 = \frac{ 7}{0.4518}[/tex]

[tex]v_2 = 15.5 \ m/s [/tex]

Generally from Bernoulli's Equation we have that

[tex]P_1 + \frac{1}{2} v_1^2 * \rho + \rho *g *h_1 = P_2 + \frac{1}{2} v_2^2 * \rho + \rho *g *h_2[/tex]

Here [tex]P_1 [\tex] is pressure in the chamber which we are to calculate , [tex]P_2 [\tex] is the atmospheric pressure with value  [tex]P_2 =  101325 \ Pa  [\tex] , [tex]v_1 [\tex] is the velocity of the water before it starts flowing with value [tex]v_1  =  0 m/s [\tex] , [tex]\rho [\tex]  is the density of water with value [tex]\rho =  1000 \ kg/m^3  [\tex]

So

       [tex]P_1  + \frac{1}{2} 0^2 * 1000 + 1000 *9.81 *0.5 = 101325  + \frac{1}{2}* 15.5^2* 1000 + 1000 *9.81 *10[/tex]  

         [tex]P_1  = 314645 \ Pa [/tex]        

Answer:

v = 5.26 10² m / s

Explanation:

We can solve this exercise using the concepts of conservation of mechanical energy, because there is no friction

starting point. Red blood cells too far away

          Em₀ = K = ½ m v²

final point. Red blood cells touching r = 8.20 10⁻⁶ m

          Em_f = U = k q₁ q₂ / r₁₂

          Em₀ = Em_f

          ½ m v² = k q₁ q₂ / r₁₂

          v = √ (2 k q₁ q₂ / m r₁₂)

we calculate

          v =√ (2 9 10⁹ 2 10⁻¹² 2.9 10⁻¹² / (4.60 10⁻¹⁴  8.20 10⁻⁶))

          v = √ (0.276775 10⁶)

          v = 0.526 10³ m / s

          v = 5.26 10² m / s

Explanation:

It is given that,

Resistance, R = 20 ohms

Voltage of the battery, V = 10 V

We can find current flowing through the circuit using Ohm's law as follows :

V = IR

[tex]I=\dfrac{V}{R}\\\\I=\dfrac{10}{20}\\\\I=0.5\ A[/tex]

It can be seen from the Ohm's law, that the current is directly proportional to the voltage. It means that if the battery is replaced by a battery that provides a larger voltage, the current through the circuit will be more than 0.5 A i.e. it increases.

The current via the circuit will be more than 0.5 A i.e. it increases.

Since

Resistance, R = 20 ohms

The voltage of the battery, V = 10 V

Now we applied the above law i.e.

V = IR

I = V/R

I = 10/20

I = 0.5 A

Based on the above calculations, we can say that the current should be directly proportional to the voltage. That means in the case when the battery should be replaced by the battery that gives the larger voltage so the current should be more than 0.5 A due to this it increased.

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Answer:

  f = 96.56 Hz

Explanation:

A soda bottle can be simulated by a system of a tube with one end open and the other closed. In this case at the closed end we have a node and at the open end a maximum, so the wavelength is

        λ = 4L / n

where n = 1, 3, 5, ...

for our case the speed of sound is

       v = λ f

we substitute

       f / v = 4L / n

calculate

      290 / (340 4) = L / n

       0.213 = L / n

In the exercise the value of n is not indicated, but we can assume that it is the fundamental answer, therefore n = 1

        L = 0.213 m

now we can calculate the frequency of the following resonance n = 3

           f = 4L v / n

           f = 4 0.213 340/3

           f = 96.56 Hz

Answer:

20 kg

Explanation:

Assuming that the board remains horizontal with the unknown mass on it, then F = kx

If we add the vertical forces to zero, we have something like this

2F - (M + 10)g = 0

2F = (M + 10)g, next, divide both sides by 2

F = (M + 10)g/2

Since we were able to sum the moments at the right end of the board to zero, we then proceed to find the unknown mass M

To start, we say, Let the clockwise moment is positive, and so

F * 4 + F * 2 - Mg * 3.5 - 10 * g * 2 = 0

4F + 2F - 3.5Mg - 10 * 2 * g = 0

6F - 3.5Mg - 10 * 2 * g = 0,

Remember from above, we say that

F = M + 10)g/2, now, all we do is substitute it inside this equation

6 * (M + 10)g/2 - 3.5Mg - 10 * 2 * g = 0

3 * (M + 10)g - 3.5Mg - 10 * g * 2 = 0, divide all sides by g(so as to eliminate it)

3 (M + 10) - 3.5M - 10 * 2= 0

3M + 30 - 3.5M - 20 = 0

-0.5M + 10 = 0

0.5M = 10

M = 10/0.5

M = 20 kg

The mass of the block that was placed on the board is; M = 20 kg

Since the two supports are the same, then it means their forces will be the same.

Thus;

Force at left support = F

Force at right support = F

Now, we are told that a block of unknown mass is placed on the stiff board. This means that the sum of the weight of the board and and the block on it will be; (M + 10)g

where M is the mass of the block.

Now, from equilibrium we know that sum of upward forces is equal to sum of downward forces. Thus;

F + F = (M + 10)g

Thus;

2F = (M + 10)g

F = ¹/₂(M + 10)g

Now, taking moments about the right end gives;

(F × 4) + (F × 2) - (M × 3.5)g - (10 × 2)g = 0

6F - 3.5Mg - 20g = 0

Put ¹/₂(M + 10)g for F to get;

6(¹/₂(M + 10)g) - 3.5Mg - 20g = 0

Divide through by g to get;

3(M + 10) - 3.5M = 20

3M + 30 - 3.5M = 20

3.5M - 3M = 30 - 20

0.5M = 10

M = 10/0.5

M = 20 kg

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Answer:

the answer is true

Explanation:

Answer:

Inertia: a tendency to do nothing or to remain unchanged; momentum:the quantity of motion of a moving body; velocity:the speed of something in a given direction; acceleration: a vehicles capacity to gain speed within a short time

Answer:

Yes the color changes. If the star is moving away from the earth the color of the star is shifted towards red (red-shift).

This is similar to the change  in sound from an approaching train whistle (higher pitch) or a receding train  whistle (lower pitch) - the Doppler Effect.

Note that the equations for the Doppler Effect for sound and for light are different - sound travels thru air but no such medium can be identified for light.

Answer:

Mass of Iron is 24.45 kg

Explanation:

Given that:

Mass of Silver, [tex]m_{S}[/tex] = 8.00 kg

Mass of Gold, [tex]m_{G}[/tex] = 15.0 kg

Initial temperature of Silver and Gold = 333 K

Initial temperature of Iron = 1737 K

Final temperature = 1337 K

Specific heat capacity of Gold-liquid, [tex]c_{G}[/tex] = 0.150 kJ/(kg K)

Specific heat of Silver-liquid, [tex]c_{S}[/tex]  = 0.280 kJ/(kg K)

Known: Specific heat capacity of Iron, [tex]c_{I}[/tex]  = 0.461 kJ/(kg K)

Therefore;

Heat lost by Iron = Heat gained by Silver + Heat gained by Gold

[tex]m_{I}[/tex] x [tex]c_{I}[/tex] x Δθ = ([tex]m_{S}[/tex] x [tex]c_{S}[/tex] + [tex]m_{G}[/tex] x [tex]c_{G}[/tex]) x Δθ

[tex]m_{I}[/tex] x 0.461 x (1737 - 1337) = (8 x 0.280 + 15 x 0.150) x (1337 - 333)

[tex]m_{I}[/tex] x 0.461 x 400 = (2.24 + 2.25) x 1004

[tex]m_{I}[/tex] x 184.4 = 4507.96

[tex]m_{I}[/tex] = [tex]\frac{4507.96}{184.4}[/tex]

[tex]m_{I}[/tex] = 24.4466

[tex]m_{I}[/tex] ≅ 24.45

The mass of Iron is 24.45 kg.

Answer:

a) 3.56 x 10^22 N

b) 3.56 x 10^22 N

Explanation:

Mass of the sun M = 2 x 10^30 kg

mass of the Earth m = 6 x 10^24 kg

Distance between the sun and the Earth R = 1.5 x 10^11 m

From Newton's law,

F = [tex]\frac{GMm}{R^2}[/tex]

where F is the gravitational force between the sun and the Earth

G is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2

m is the mass of the Earth

M is the mass of the sun

R is the distance between the sun and the Earth.

Substituting values, we have

F = [tex]\frac{6.67*10^{-11}*2*10^{30}*6*10^{24}}{(1.5*10^{11})^2}[/tex] = 3.56 x 10^22 N

A) The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to 3.56 x 10^22 N

b) The force exerted by the Earth on the Sun = 3.56 x 10^22 N

Part A:

The magnitude of the gravitational force exerted by the Sun on the Earth is :

From Newton's law,

F = GmM/F

Substituting values,

F =  = 3.56 x 10^22 N

The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to 3.56 x 10^22 N.

Part B:

The force exerted by the Earth on the Sun = 3.56 x 10^22 N.

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Answer:

a) surface   λ = 5 10⁻⁷ m

   nuecleus λ=  2,890 10⁻⁹ m

there is a mixture of wavelengths giving rise to white light

b) True color sun is White

Explanation:

Wien's law establishes a relationship between the temperature of a star and the maximum wavelength emitted

          λ T = 2,898 10⁻³

if we calculate the wavelength of the Sun

T = 5800K       λ = 5 10⁻⁷ m = 5000 nm

this is the surface temperature

in the part of the  internal Sunspart (Nucleus)  about 10⁶ degrees this is where the nuclear reaction occurs

in this region the photons are very short λ

      λ = 2,890 10⁻³ / 106

      λ=  2,890 10⁻⁹ m

these photons are absorbed and emitted on their way to the surface of the sun many times, therefore there is a mixture of wavelengths giving rise to white light

b) Sunlight when it reaches Earth is absorbed by atmospheric gases, mainly Blue, as it is absorbed by Nitrogen, it lets through mainly red and red

True color sun is White

Answer:

[tex]\Rightarrow L=70 .03472 cm[/tex]

Explanation:

For convenience, let's represent the thermal expansion coefficient by [tex]\alpha[/tex], i.e. [tex]????=\alpha[/tex].

Given that, for steel [tex]\alpha =1.24\times 10^{-5}[/tex] °[tex]C^{-1}[/tex],

initial length, [tex]L_0=70 cm[/tex], initial temperature, [tex]T_0=70[/tex] °[tex]C[/tex], and the final temperature, [tex]T=110[/tex] °[tex]C[/tex].

Let the length of the rod at [tex]T=110[/tex] °[tex]C[/tex] be [tex]L[/tex].

Now, change in length, [tex]\Delta L=\alpha L_0 \Delta T[/tex]

[tex]\Rightarrow \Delta L=\alpha L_0 (T-T_0)[/tex]

[tex]\Rightarrow L-L_0=1.24\times 10^{-5}\times 70 (110-70)[/tex]

[tex]\Rightarrow L-70=1.24\times 10^{-5}\times 70 \times 40[/tex]

[tex]\Rightarrow L=70 + 0.03472 cm[/tex]

[tex]\Rightarrow L=70 .03472 cm[/tex]

Hence, the length of the rod at [tex]T=110[/tex] °[tex]C[/tex] be [tex]70.03472 cm[/tex].

Answer:

its 667m just took the test

Explanation:

Answer:

he factor for the temporal part 1.296 107 s² = h²

 m / s² = 12960 km / h²

Explanation:

This is a unit conversion exercise.

In the unit conversion, the size of the object is not changed, only the value with respect to which it is measured is changed, for this reason in the conversion the amount that is in parentheses must be worth one.

In this case, it is requested to convert a measure km/h²

Unfortunately, it is not clearly indicated what measure it is, but the most used unit in physics is   m / s² , which is a measure of acceleration. Let's cut this down

the factor for the distance is 1000 m = 1 km

the factor for time is 3600 s = 1 h

let's make the conversion

        m / s² (1km / 1000 m) (3600 s / 1h)²

note that as time is squared the conversion factor is also squared

        m / s² = 12960 km / h²

the factor for the temporal part 1.29 107 s² = h²

Answer:

it can be cross pollinated as well as self pollinated

it has short life style

pea plant has many contrasting character in pair example tall, short.

large number of offspring are produced from hybrid plants

Answer:

ΔV = 0.816 L

Explanation:

The change in volume of the fluid upon heating is given by the following formula:

ΔV = βVΔT

where,

ΔV = Increase in Volume of Fluid = Volume of Overflow = ?

β = Coefficient of volumetric expansion of fluid = 600 x 10⁻⁶ °C⁻¹

ΔT = Change in Temperature = Final Temperature - Initial Temperature

ΔT = 95°C - 10°C = 85°C

Therefore,

ΔV = (600 x 10⁻⁶ °C⁻¹)(16 L)(85° C)

ΔV = 0.816 L

Answer:

4 s

Explanation:

Given:

Δx = 12 m

v₀ = 6 m/s

v = 0 m/s

Find: t

Δx = ½ (v + v₀) t

12 m = ½ (0 m/s + 6 m/s) t

t = 4 s

Answer:

The value is [tex]V =168.75\ millivolt [/tex]

Explanation:

From the question we are told that

The workdone is [tex]W= 2.70 * 10^{-20 } \ J[/tex]

Generally charge on the positive sodium ion is equivalent to the charge on a proton, the value is [tex]e = 1.60 *10^{-19} \ C[/tex]

Generally the potential difference between the inner and outer surfaces of the cell is mathematically represented as

[tex]V = \frac{W}{e}[/tex]

=>      [tex]V  =  \frac{2.70 * 10^{-20 } }{1.60 *10^{-19} }[/tex]

=>     [tex]V  = 0.16875 \  V[/tex]

converting to millivolt

         [tex]V  = 0.16875 * 1000 [/tex]

    =>   [tex]V  =168.75\ millivolt [/tex]

the answer is c because you have to move the decimal back because of the negative

Answer:

Acceleration Remains Constant and Velocity decreases.

Explanation:

Answer:

Explanation:

M= 0.003 kg

F=300N

a=?

F=mass*acceleration

a=F/m

a=300/0.003

a=100,000m/s^2

Answer:

Explanation:

A vector is any substance that has both magnitude and direction. lndeed a vector quantity has both magnitude and direction. Example of a vector quantity is velocity. Determining the velocity of an object allows one to determine how fast and in what direction the object moves. Velocity also affords the opportunity to compare the motion of more than one vectors.

Answer:

The final temperature is [tex]\left(T_1^{C_1}+T_2^{C_2}\right)^{\frac {1}{C_1+C_2}}[/tex]

and the maximum amount of workdone is [tex]C_1T_1 + C_2T_2-(C_1+C_2)\left(T_1^{C_1}+T_2^{C_2}\right)^{\frac {1}{C_1+C_2}}[/tex].

Explanation:

Assume that [tex]R_1[/tex] is the reservior having temperature [tex]T_1 K[/tex] and heat capicity [tex]C_1 \frac JK[/tex] and [tex]R_2[/tex] is the reservior having temperature [tex]T_2[/tex] and heat capicity [tex]C_2 \frac JK[/tex].

The work will be extracted till that both the reservior reach the thermal equilibrium. Let the final temperature of both the reservior is [tex]T_f[/tex].

Let total [tex]Q_1[/tex] heat is extracted by the heat engine from the reservior [tex]R_1[/tex] and its temperature decreases from [tex]T_1[/tex] to [tex]T_f[/tex] and [tex]Q_2[/tex] heat is rejected by the heat engine to the reservior [tex]R_2[/tex] and its temperature decreases from [tex]T_2[/tex] to [tex]T_f[/tex].

So, The maximum amount of work done,

[tex]w= Q_1 - Q_2\; \cdots (i)[/tex]

Now, as the heat engine is reversible, so change is entropy for the universe is 0, which means sum of change in entropy for the ststem as well as surrounding is 0.

As shown in figure, the system is the reversible engine, so, change is entropy for the system is 0. Hence, change in entropy for the the surrounding is 0.

As temperature of [tex]R_1[/tex] is changing fron [tex]T_1[/tex] to [tex]T_f[/tex], so, change in entropy of surrounding due to transfer of [tex]Q_1[/tex] is [tex]C_1 \ln \frac {T_f}{T_1}[/tex].

Similarly,  change in entropy of surrounding due to transfer of [tex]Q_2[/tex] is [tex]C_2 \ln \frac {T_f}{T_2}[/tex].

As the net change in entropy of the surrounding is 0.

[tex]\Rightarrow C_1 \ln \frac {T_f}{T_1}+C_2 \ln \frac {T_f}{T_2}=0[/tex]

[tex]\Rightarrow \ln \left( \frac {T_f}{T_1} \right)^{C_1}+ \ln \left( \frac {T_f}{T_2}\right)^{C_2}=0[/tex]

[tex]\Rightarrow \ln \left(\frac {T_f}{T_1}\right)^{C_1}=- \ln \left( \frac {T_f}{T_2}\right)^{C_2}[/tex]

[tex]\Rightarrow \ln \left(\frac {T_f}{T_1}\right)^{C_1}= \ln \left( \frac {T_2}{T_f}\right)^{C_2}[/tex]

[tex]\Rightarrow \left( \frac {T_f}{T_1}\right)^{C_1}=\left( \frac {T_2}{T_f}\right)^{C_2}[/tex][taking anti-log both the sides]

[tex]\Rightarrow T_f^{(C_1 +C_2)}=T_1^{C_1}+T_2^{C_2}[/tex]

[tex]\Rightarrow T_f=\left(T_1^{C_1}+T_2^{C_2}\right)^{\frac {1}{C_1+C_2}}\; \cdots (ii)[/tex]

This is the required final temperature.

Now, from equarion (i), the maximum amount of work done is

[tex]w= Q_1 - Q_2[/tex]

As [tex]Q=C\Delta T[/tex]

[tex]\Rightarrow w=C_1(T_1-T_f)-C_2(T_f-T_2)[/tex]

[tex]\Rightarrow w=C_1T_1 + C_2T_2-(C_1+C_2)T_f[/tex]

From equation [tex](ii)[/tex],

[tex]w=C_1T_1 + C_2T_2-(C_1+C_2)\left(T_1^{C_1}+T_2^{C_2}\right)^{\frac {1}{C_1+C_2}}[/tex]

This is the required maximum workdone.

Answer:

The time taken is [tex]t = 52.5 \ s [/tex]

Explanation:

From the question we are told that

The speed limit is [tex]v__{{l}}} = 96.0 \ km/hr = \frac{96 * 1000}{3600} = 26.7 \ m/s[/tex]

The velocity of the motorist is [tex]v_m = 107 \ km/hr = \frac{107 * 1000}{3600} = 29.72 \ m/s[/tex]

The chase speed of the motorcycle patrolman is [tex]v = 131 \ km/hr = \frac{131 *1000}{3600} = 36.39 \ m/s[/tex]

The relative distance between the motorcycle patrolman and the speeder is d= 350 m

Generally the relative speed between the the motorcycle patrolman and the speeder is mathematically represented as

[tex]v_r = v - v_m[/tex]

=> [tex]v_r = 36.39 - 29.72[/tex]

=> [tex]v_r = 6.67 \ m/s [/tex]

Generally the time taken is mathematically represented as

[tex]t = \frac{v_r}{d}[/tex]

=>     [tex]t =  \frac{350}{ 6.67}[/tex]

=>    [tex]t =  52.5 \  s [/tex]

Answer:

groups

Explanation:

I got a 100 on my quiz

Answer:

It's groups

Explanation:

2020 quiz

Answer:

temperature

Explanation:

when you put a water on the stove the water will start to boil there for temperature