is power a vector quantity​

Answer:

I like to memorize excerpt from articles to solve and answer questions like these. I hope this can help, it's from study.com: "The relationship between voltage, current, and resistance is described by Ohm's law. This equation, i = v/r, tells us that the current, i, flowing through a circuit is directly proportional to the voltage, v, and inversely proportional to the resistance, r."

We are given:

extension of the spring (x) = 10 cm OR 0.1 m

Force applied to keep it in position (F) = 500 N

Solving for the spring constant:

We know that F = kx (when the spring is extended)

replacing the variables with the given values

500 = k * 0.1

k = 5000 N/m

This means that to extend the spring by 1 m, we have to apply a force of 5000 N

Answer:

(a) 0.993 s

(b) 14.0 N/m

(c) -3.02 m/s

(d) -6.01 m/s²

Explanation:

(a) The block's position can be modeled as a cosine wave:

x(t) = A cos(ωt)

where A is the amplitude (in this case, 50 cm) and ω is the angular frequency.

At t = 0.200 s, x(t) = 15.0 cm.

15.0 cm = 50.0 cm cos((0.200 s) ω)

0.3 = cos((0.2 s) ω)

1.266 rad = (0.2 s) ω

ω = 6.33 rad/s

The period is:

T = (2π rad) (1 s / 6.33 rad)

T = 0.993

(b) For a spring-mass system, ω = √(k/m).  The mass of the block is 0.350 kg, so:

ω = √(k/m)

6.33 rad/s = √(k / 0.350 kg)

6.33 rad/s = √(k / 0.350 kg)

40.1 rad/s² = k / 0.350 kg

k = 14.0 N/m

(c) Energy is conserved:

EE₀ = EE + KE

½ kx₀² = ½ kx² + ½ mv²

kx₀² = kx² + mv²

(14.0 N/m) (0.50 m)² = (14.0 N/m) (0.15 m)² + (0.35 kg) v²

v = -3.02 m/s

Alternatively, we can take the derivative of our position equation:

v(t) = -Aω sin(ωt)

v = -(0.50 m) (6.33 rad/s) sin((6.33 rad/s) (0.2 s))

v = -3.02 m/s

(d) Sum of forces on the block:

∑F = ma

-kx = ma

a = -kx / m

a = -(14.0 N/m) (0.15 m) / (0.350 kg)

a = -6.01 m/s²

Alternatively, we can take the derivative of our velocity equation:

a(t) = -Aω² cos(ωt)

a = -(0.50 m) (6.33 rad/s)² cos((6.33 rad/s) (0.2 s))

a = -6.01 m/s²

Answer:

0.130

Explanation:

From the given data, the coefficient of static friction for each trial are:

1. 0.053

2. 0.081

3. 0.118

4. 0.149

5. 0.180

6. 0.198

The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

                                              = 0.779

So that;

the average coefficient of static friction = [tex]\frac{sum of coefficient of static friction}{number of trials}[/tex]

                                              = [tex]\frac{0.779}{6}[/tex]

                                              = 0.12983

The average coefficient of static friction is 0.130

The average coefficient of static friction is 0.13.

The coefficient of static friction is obtained using the formula; μ = F/R

Where;

F = force acting on the body

R = reaction

μ = coefficient of static friction

The average of measurements is given as; ∑summation of measurements/number of measurements

We can see from the question that there were 6 measurements of the coefficient of static friction. Hence, the average coefficient of static friction is obtained from;

0.053 + 0.081 + 0.118 + 0.149 + 0.180 +  0.198/6

= 0.13

The average  coefficient of static friction is 0.13

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Answer:

Explanation:

Frequency is oscillations per second.

So we have to find the Number of seconds she paced.

2 minutes = 2 X 60

               = 120 seconds

Therefore,

Her frequency = 10 / 120

                       = 1/12 Hertz

Answer:

Yes

Explanation:

I think this is because when you go out of the room and going to a hotter room you then get the heat from that room. It then becomes warmer in the room you are coming from because your body got the heat from the outside the room. I think it is because of body temperature.

HOPE THIS HELPED

Answer:

42

Explanation:

p=mv

momentum=mass*velocity

x=2.8*15

x=42

Explanation:

so sorry

don't know but please mark me as brainliest please

Given :

Pressure, P = 0.4 lb/in².

Depth, h = 42 in.

To Find :

The density of a liquid.

Solution :

Pressure at height h of a liquid of density [tex]\rho[/tex] is given by :

[tex]P = \rho g h[/tex]   ....1)

Here, g = 386.04 in/s²( acceleration due to gravity )

Putting all values in equation 1, we get :

[tex]P = \rho g h\\\\0.4 = \rho\times 386.04 \times 42\\\\\rho=\dfrac{0.4}{42\times 386.04}\ lb/in^3\\\\\rho=\dfrac{0.4}{42\times 386.04}\times 12^3\ lb/ft^3\\\\\rho=0.0426\ lb/ft^3[/tex]

Hence, this is the required solution.

Answer:

a i belive

Explanation:

the univerce is VERY large so a, if im wrong i apologise :(

The unit that should be used to measure energy when calculating specific heat capacity is Energy is in Joules ( J ).

The specific heat capacity should be determined in joules per kilogram degree-celsius ( J k g − 1 ∘ C − 1 ).

It is to be supplied for the substance with respect to mass and it increased the temperature.

Hence, The unit that should be used to measure energy when calculating specific heat capacity is Energy is in Joules ( J ).

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The unit used to measure energy when calculating specific heat capacity is the joule (J).

The joule (J), a unit used to quantify energy, is used to calculate specific heat capacity. In the International System of Units (SI), the joule serves as the default unit of energy.

It is described as the quantity of energy that is delivered when one newton of force is exerted across a one-meter distance.

Thus, the quantity of heat energy needed to increase the temperature of a particular substance by a specific amount is measured as specific heat capacity. J/kg°C, or joules per kilogramme per degree Celsius, is the unit of measurement.

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Answer:

[tex]5\sqrt{2} \ N[/tex]

Explanation:

It is given that,

A 5-newton force directed east and a 5-newton force directed  north.

We need to find the resultant of the two forces.

As the two forces are acting in perpendicular to each other. The resultant of two such forces is given by :

[tex]F=\sqrt{5^2+5^2} \\\\=5\sqrt{2} \ N[/tex]

So, the resulatnt force is [tex]5\sqrt{2} \ N[/tex]

Answer:

L/x = 3.5 (The ratio of length of beam to the distance from heavier ball to pivot).

Explanation:

In order for the system to be in equilibrium, the moment created by both masses about the pivot point must be equal:

m₁x = m₂y

where,

m₁ = 5 kg

m₂ = 2 kg

x = distance of 5 kg ball from pivot

y = distance of 2 kg ball from pivot

Therefore,

(5 kg)x = (2 kg)y

y = (5kg/2kg)x

y = 2.5 x

but,

x + y = L

where,

L = length of beam

using the value of y from the previous equation:

x + 2.5 x = L

3.5 x = L

L/x = 3.5 (The ratio of length of beam to the distance from heavier ball to pivot).

Answer:

The mass of the cart is 150 kg.

Explanation:

Given that,

Mass of a boy, m₁ = 50 kg

Initial speed of boy, u₁ = 10 m/s

Initial speed of car, u₂ = 0 (at rest)

The speed of the cart with the boy on it is 2.50 m/s, V = 2.5 m/s

Let m₂ is the mass of the cart. Using the conservation of momentum as follows :

[tex]m_1u_1+m_2u_2=(m_1+m_2)V\\\\50(10)+m_2(0)=(50+m_2)(2.5)\\\\500=125+2.5m_2\\\\375=2.5m_2\\\\m_2=150\ kg[/tex]

So, the mass of the cart is 150 kg.

Answer:

Explanation:

a) centripetal acceleration is the acceleration of a body in a circular path. It is expressed as;

a = mv²/r

m is the mass of proton = 1.67×10^−27kg

v is the velocity = 3.00×10^8m/s

r is the radius

Since C = 2πr

7000m = 2πr

r = 7000/2π

r = 1114.08m

Substitute

a = 1.67×10^−27 (3.00×10^8)²/1114.08

a =  1.67×10^−27 * 9×10^16/1114.08

a = 15.03*10^-11/1114.08

a = 0.001346*10^-11

a = 1.346*10^-14m/s²

b) Force on the proton = mass * acceleration

Force = 1.67×10^−27kg * 1.346*10^-14

Force = 2.246*10^-41N

hence the force on the proton is 2.246*10^-41N

Answer:

It was not fired from the client's gun because the chair slid only 3 centimeters . If it had been fired from the client's gun the chair would slid 25.82 centimeters.

Explanation:

According to the law of conservation of momentum the momentum of the system before collision must be equal to the momentum of the system after the collision.

M1u1= m2u2

Let M1 = mass of the chair = 20kg

     m2= mass of the bullet= 10g= 0.001kg

      u1= velocity of the chair before collision = zero m/s

      u2 =  velocity of the bullet before collision = zero m/s

         v1= velocity of the chair after collision = ?  m/s  

        v2 =  velocity of the bullet after collision = 450 m/s

After collision their velocities change from u1 to v1 and u2 to v2 so

M1v1= m2v2

v1= m2v2/M1

v1= 0.01 *450/ 20=  0.225 m/s

Now according to the law of conservation of energy the energy of the system before collision must be equal to the energy of the system after the collision.

The energy of the chair after the bullet is hit is

KE of the chair + KE of the bullet=  1/2 (M)(v1)²+ 1/2 m(v2)²=

1/2 ( 20) (0.225 )² + 1/2 (0.01) (450)²

= 0.50625 + 1012.5=  1013.00625 Joules

Frictional force = Coefficient of kinetic force of wood on wood ( M+m) g

                           = 0.2* ( 20.01) 9.8=   39.2196 N

Work done by friction = frictional force * distance

If law of conservation of energy is applied  the KE  must be equal to the work done

KE = W

W= f*d

KE= F*d

d = KE/f= 1013.00625/ 39.2196= 25.82 cm

The chair did not move 25.82 cm .

It only moved 3 centimeter.

Hence the bullet fired was not from the client's gun.

Answer:

Magnitude of gravitational force between this planet and its moon: approximately [tex]1.37 \times 10^{20}\; \rm N[/tex].

Acceleration of this moon towards the planet: approximately [tex]5.49 \times 10^{-3}\; \rm m \cdot s^{-2}[/tex].

Acceleration of this planet towards its moon: approximately [tex]2.29 \times 10^{-5}\; \rm m \cdot s^{-2}[/tex].

Explanation:

Look up the gravitational constant, [tex]G[/tex]:

[tex]G \approx 6.67 \times 10^{-11}\; \rm m^3\cdot kg^{-1} \cdot s^{-2}[/tex].

Assume that both this planet and its moon are spheres of uniform density. When studying the gravitational interaction between this planet and its moon, this assumption allows them to be considered as two point masses.

The formula for the size of gravitational force between two point masses [tex]m_1[/tex] and [tex]m_2[/tex] with a distance of [tex]r[/tex] in between is:

[tex]\displaystyle F = \frac{G \cdot m_1 \cdot m_2}{r^2}[/tex],

where [tex]G[/tex] is the gravitational constant.

Let [tex]m_1[/tex] and [tex]m_2[/tex] denote the mass of this planet and its moon, respectively.

Calculate the size of gravitational force between this planet and its moon:

[tex]\begin{aligned} F &= \frac{G \cdot m_1 \cdot m_2}{r^2} \\ &\approx \frac{6.67 \times 10^{-11}\; \rm m^3 \cdot kg^{-1} \cdot s^{-2} \times 6.00 \times 10^{24}\; \rm kg \times 2.50 \times 10^{22}\; \rm kg}{{\left(2.70 \times 10^{8}\; \rm m\right)}^2} \\ &\approx 1.37 \times 10^{20}\; \rm N\end{aligned}[/tex].

Assume that other than the gravitational force between this planet and its moon, all other forces (e.g., gravitational force between this planet and the star) are negligible. The magnitude of the net force on the planet and on the moon should both be approximately [tex]1.37 \times 10^{20}\; \rm N[/tex].

Apply Newton's Second Law of motion to find the acceleration of this planet and its moon:

[tex]\displaystyle \text{acceleration} = \frac{\text{net force}}{\text{mass}}[/tex].

For this moon:

[tex]\displaystyle \frac{1.37 \times 10^{20}\; \rm N}{2.50 \times 10^{22}\; \rm kg} \approx 5.49\times 10^{-3}\; \rm m \cdot s^{-2}[/tex].

For this planet:

[tex]\displaystyle \frac{1.37 \times 10^{20}\; \rm N}{6.00 \times 10^{24}\; \rm kg} \approx 2.29 \times 10^{-5}\; \rm m \cdot s^{-2}[/tex].

Answer:

840 m/s

Explanation:

Given that,

In the first second the rocket ejects 1/160 of its mass as exhaust gas and has an acceleration of 14.0 m/s².

We need to find the speed of the exhaust gas relative to the rocket.

The thrust of rocket is given by :

[tex]T=v_{gas}\dfrac{dm}{dt}\\\\ma=v_{gas}\dfrac{dm}{dt}\\\\v_{gas}=\dfrac{ma}{\dfrac{dm}{dt}}\\\\v_{gas}=\dfrac{14m}{\dfrac{1}{60}m}\\\\v_{gas}=840\ m/s[/tex]

So, the speed of the exhaust gas relative to the rocket is 840 m/s.

The speed ( Vgas) of the exhaust gas relative to rocket is : 840 m/s

Given data :

In first round Rocket ejects  1/60 of mass as exhaust gas

Acceleration of rocket ( a ) = 14.0 m/s²

Determine the speed of the exhaust gas relative to rocket

We will apply the equation for Rocket thrust

T = Vgas * [tex]\frac{dm}{dt}[/tex]

where : T = ma

∴ Vgas = ma / [tex]\frac{dm}{dt}[/tex]

            = 14 m / [tex]\frac{1}{60}[/tex] m

therefore V gas = 840 m/s

Hence we can conclude that the  speed ( Vgas) of the exhaust gas relative to rocket is : 840 m/s

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Answer:

Explanation:

distance = 1.25 mile

time traveled = 5 min.

find velocity in miles / hour

solution:

use the formula: velocity = distance / time

velocity =     1.25 mile     x 60 min  

                     5 min            1 hour

velocity = 15 miles / hour

Answer:

so what is your question that's not a question

An asteroid is a minor planet in the inner solar system. The asteroid orbits the sun and millions of asteroids exit as remains of planetesimals.

Hence there is no contact as no gravitational force.

Learn more about the astronaut standing on the surface of a spherical asteroid.

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Answer:

I think they are called balanced forces

Explanation:

Answer:

Star Y is showing radial motion

Star Y is moving toward the Earth

Explanation:

Just answered this question on a quiz and got it right.

Star Y is showing radial motion

Star Y is moving towards earth.

Spectrum is defined as the arrangement of radiations emitted by atoms or molecules based on their characteristic wavelength and frequency.

Here,

The spectrum of the star Y is given and is compared to a reference hydrogen spectrum. The range of wavelengths of the star is given. The spectrum of the star shows that the wavelengths are shifted such that from longer wavelengths to shorter wavelength. This phenomenon of shift in wavelengths from higher to lower is known as blue shift. The star shows blue shift that means shifted to shorter wavelength or it can be said as shifted from lower frequency to higher frequency. As a result, it can be concluded that the star Y is moving towards the earth which implies Star Y is showing radial motion.

Hence,

The star Y is showing radial motion.

Star Y is moving towards the earth.

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Answer:

Option (c) is correct.

Explanation:

Acceleration of an object is given by the formula as follows :

[tex]a=\dfrac{v-u}{t}[/tex]

Where

u and v are initial and final velocity

t is time

(v-u) is also called the change in velocity

So, the acceleration of an object is equal to the rate of change of velocity. Hence, the correct option is (c) " Change in its velocity divided by the change in time".

Answer:

20 K

Explanation:

It is given that,

The change in temperature is 20 C.

We need to find the change in thermodynamic temperature.

If teperauture T₁ = 0° C = 0+273 = 273 K

T₂ = 20° C = 20 + 273 = 293 K

The change in temperature,

[tex]\Delta T=T_2-T_1\\\\=293-273\\\\=20\ K[/tex]

So, the change in temperature of 20°C is equivalent to 20 K.

The change in thermodynamic temperature is equal to 20 Kelvin.

Given the following data:

To determine the change in thermodynamic temperature:

A thermodynamic temperature can be defined as an absolute measure of the average total internal energy possessed by a body or an object. Thus, thermodynamic temperature is typically measured in Kelvin (K).

Mathematically, the change in temperature of an object is given by the formula:

[tex]\theta = T_f - T_i[/tex]

Where:

Next, we would convert the values in degree Celsius to Kelvin:

Conversion:

0°C = 273 K

20°C = 273 + 20 = 293 K

For the change in thermodynamic temperature:

[tex]\theta = 293 - 273\\\\\theta =20\;K[/tex]

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Answer:

1.63 m

Explanation:

The following data were obtained from the question given above:

Pressure (P) = 120 mmHg

Density (d) = 1000 Kg/m³

Height (h) =?

Next, we shall convert 120 mmHg to N/m². This can be obtained as follow:

760 mmHg = 101325 N/m²

Therefore,

120 mmHg = 120 mmHg × 101325 N/m² / 760 mmHg

120 mmHg = 15998.68 N/m²

Thus, 120 mmHg is equivalent to 15998.68 N/m².

Finally, we shall determine the height to which the IV bag should be placed so that the fluid can enter the blood stream. This is illustrated below:

Pressure (P) = 15998.68 N/m².

Density (d) = 1000 Kg/m³

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) =?

P = dgh

15998.68 = 1000 × 9.8 × h

15998.68 = 9800 × h

Divide both side by 9800

h = 15998.68 / 9800

h = 1.63 m

Therefore, the the IV bag should be placed at a height of 1.63 m

Answer:

7 because salt lake and Southis weat

Answer:

The minimum downward force on the box that will keep it from slipping is 63.14 N

Explanation:

Given;

mass of the object, m = 30 kg

applied force, f = 125 N

coefficient of static friction, μ = 0.35

Normal reaction (R) is acting upwards, weight of the box (mg) is acting downwards and the minimum downward force (F) on the box that will keep it from slipping is also acting downwards.

The net vertical forces on the box is given by;

R - mg - F = 0

F = R - mg

Now, determine normal reaction, R

f = μR

R = f / μ

R = 125 / 0.35

R = 357.14 N

Finally, determine the minimum downward force on the box that will keep it from slipping;

F = R - mg

F = 357.14 - (30 x 9.8)

F = 357.14 - 294

F = 63.14 N

Therefore, the minimum downward force on the box that will keep it from slipping is 63.14 N

Answer:

Explanation:

To get the focal length, we will use the lens formula;

1/f = 1/u + 1/v

f is the focal length

u is the object distance

v is the image distance

Given

since the physicist's left eye is myopic, it will be corrected using concave lens and the image distance is negative.

u = 35cm

v = -2.0

1/f = 1/35-1/2

1/f = 2-35/70

1/f = -33/70

f = -70/33

f = -2.12 cm

f = -0.0212m

Power of a lend is the reciprocal of its focal length

Power of the lens = 1/f

P = 1/-0.0212

P = -47.17dioptres

The power of the lens is -47.17D