are Car Travels certain distance with a speed of 50 kilometre per hour and returns with a speed of 40 kilometre per hour what is the average speed for the whole journey
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Answer:

so 9/3=3 current is 3 amperes

Explanation:

The fomula to calculate resistance is:

voltage/cutrent

9 V/3 A= 3 ohms

Answer:

This plasma mostly consists of electrons

Answer:

a)  3.0  10⁴ m / s, b) 3.7 10¹ m, c) 0.750 kg, d) 4 10¹² m²,  e)  75 m, f) 60 m²

g) 5.207 10³ m², e) 4.847 10⁷ s

Explanation:

The international system (SI) of measurements has as fundamental units the meter for length, the second for time and kilogram for mass.

Let's reduce the different magnitudes to the SI system

a) 30 km / h (1000m / 1 km) (1 h / 3600 s) = 3.0 10⁴ m / s

b) 37 Dm (10 m / 1 Dm) = 3.7 10¹ m

c) 750 g (1 kg / 10,000 g) = 0.750 kg

d) 4 10⁶ km² (1000 m / 1km) ² = 4 10¹² m²

e) 7500 cm (1 m / 100 cm) = 75 m

f) 600000 cm² (1m / 10² cm) ² = 60 m²

g) 520700000 mm³ (1 m / 10³ mm) ³ = 5.20700000 109/10 ^ 6

  = 5.207 10³ m²

e) 3.4 years (l65 days / 1 yr) (24 h / 1 day) (3600 s / 1h) = 4.847 10⁷ s

Answer:

Thus, the time for the first lamp is 44 minutes.

Explanation:

Power of first lamp, P' = 1000 W

Power of second lamp, P'' = 4400 W

time for second lamp, t'' = 10 minutes

Let the time for first lamp is t'.

As the energy is same, so,

P' x t' = P'' x t''

1000 x t' = 4400 x 10

t' = 44 minutes

Answer:

The second one.

Explanation:

It caused both to change speed because they have both the same mass.

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

[tex]I = m\cdot (\vec{v}_{2} - \vec{v_{1}})[/tex] (1)

Where:

[tex]I[/tex] - Impulse, in kilogram-meters per second.

[tex]m[/tex] - Mass, in kilograms.

[tex]\vec{v_{1}}[/tex] - Initial velocity of the hockey park, in meters per second.

[tex]\vec{v_{2}}[/tex] - Final velocity of the hockey park, in meters per second.

If we know that [tex]m = 0.2\,kg[/tex], [tex]\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right][/tex] and [tex]\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right][/tex], then the impulse applied by the stick to the park is approximately:

[tex]I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right][/tex]

[tex]I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right][/tex]

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

Answer:

0.893 rad/s in the clockwise direction

Explanation:

From the law of conservation of angular momentum,

angular momentum before impact = angular momentum after impact

L₁ = L₂

L₁ = angular momentum of bullet = + 9 kgm²/s (it is positive since the bullet tends to rotate in a clockwise direction from left to right)

L₂ = angular momentum of cylinder and angular momentum of bullet after collision.

L₂ = (I₁ + I₂)ω where I₁ = rotational inertia of cylinder = 1/2MR² where M = mass of cylinder = 5 kg and R = radius of cylinder = 2 m, I₂ = rotational inertia of bullet about axis of cylinder after collision = mR² where m = mass of bullet = 0.02 kg and R = radius of cylinder = 2m and ω = angular velocity of system after collision

So,

L₁ = L₂

L₁ = (I₁ + I₂)ω

ω = L₁/(I₁ + I₂)

ω = L₁/(1/2MR² + mR²)

ω = L₁/(1/2M + m)R²

substituting the values of the variables into the equation, we have

ω = L₁/(1/2M + m)R²

ω = + 9 kgm²/s/(1/2 × 5 kg + 0.02 kg)(2 m)²

ω = + 9 kgm²/s/(2.5 kg + 0.02 kg)(4 m²)

ω = + 9 kgm²/s/(2.52 kg)(4 m²)

ω = +9 kgm²/s/10.08 kgm²

ω = + 0.893 rad/s

The angular velocity of the cylinder bullet system is 0.893 rad/s in the clockwise direction-since it is positive.

Answer:

Explanation:

F = ma is a linear equation. This means that the Force change as the accleration changes. And vice versa. If the Force is cut in thirds, then the acceleration is also cut in thirds. Let's do some math on this just to prove it, shall we?

We know that at first, the F = 15. Let's give this object a mass of 5kg. That means that

15 = 5a so

a = 3

Then the F is cut into thirds, so

5 = 5a so

a = 1

The second acceleration is one-third of the first one, where the Force is 3 times greater.

Answer:

Slow-moving rivers generate extensive floodplains and meanders through erosion and deposition. Stream and river deposition can result in the formation of alluvial fans and deltas. Natural levees may be formed by floodwaters. Caves and sinkholes can arise as a result of groundwater erosion and deposition.

Explanation:

s

The erosion and deposition of sediments in a river creates broad floodplains and meanders.

Deposition occurs when sediment, a combination of soil and rock produced by weathering, is eroded and transported to a new area.

Deposition is the act of depositing silt that has been transported by the wind, water, sea, or ice.

Earthen materials are worn away during erosion, a geological process in which they are moved by wind or water.

The removal of soil, rock, or dissolved material from one area on the Earth's crust and subsequent transport to another region for deposition are known as erosional processes. Erosion differs from weathering, which is a static process.

Given data ,

Let the erosion and deposition of sediments be deposited in a river

Now , Pebbles, sand, mud, and salts that have been dissolved in water can all be used to convey sediment. Afterwards, salts may be left behind by organic action.

Now , floodplain is caused when erosion happens

And , A floodplain is a broad, level or nearly level area of land where the stream flows.

Meandering streams that wander from side to side broaden the plain by eroding it during the formation of the plain. Flooding can occur when stream flows overflow from their channel due to very excessive rainfall or quick snow-melt.

Hence , floodplain and meandering occurs due to erosion and deposition

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The complete question is as follows: A student is subjected to a reaction force of 10 N northward from a 5 kg block while pushing the block over a smooth, level surface. Ignoring friction, what is the acceleration of the block?

Answer: The acceleration of the block is [tex]2 m/s^{2}[/tex].

Explanation:

Given: Force = 10 N

Mass = 5 kg

It is known that force applied on an object is the product of mass and acceleration.

Mathematically, [tex]F = m \times a[/tex]

where,

F = force

m = mass

a = acceleration

Substitute the values into above formula as follows.

[tex]F = m \times a\\10 N = 5 kg \times a\\a = \frac{10}{5}\\= 2 m/s^{2}[/tex]

Thus, we can conclude that the acceleration of block is [tex]2 m/s^{2}[/tex].

Answer:

F = ⅔ F₀

Explanation:

For this exercise we use Coulomb's law

         F = k q₁q₂ / r²

let's use the subscript "o" for the initial conditions

          F₀ = k q² / r²

now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r

we substitute

          F = k 4q² / 9 r²

          F = k q² r² 4/9

          F = ⅔ F₀

Answer:

Did you try searching it up

Explanation:

Answer:

P = V * Q      potential energy = potential * charge

V = =3.09 J / 6.93 * 10E-4 C = 4460 Joules / Coulomb

The electric potential, V at the point given the data from the question is –4458.87 V

The electric potential or electromotive force (EMF) is defined as the energy supplied by a battery per unit charge. Mathematically, it can be expressed as:

Electromotive force (EMF) = Work (W) / charge (Q)

V = EMF = W / Q

V = W / Q

V = –3.09 / 6.93×10⁻⁴

V = –4458.87 V

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Answer:

[tex]W=17085KJ[/tex]

Explanation:

From the question we are told that:

Height [tex]H=16m[/tex]

Radius [tex]R=3[/tex]

Height of water [tex]H_w=9m[/tex]

Gravity [tex]g=9.8m/s[/tex]

Density of water [tex]\rho=1000kg/m^3[/tex]

Generally the equation for Volume of water is mathematically given by

 [tex]dv=\pi*r^2dy[/tex]

 [tex]dv=\frac{\piR^2}{H^2}(H-y)^2dy[/tex]

Where

   y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

 [tex]dw=(pdv)g (H-y)[/tex]

Substituting dv

 [tex]dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)[/tex]

 [tex]dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy[/tex]

Therefore

 [tex]W=\int dw[/tex]

 [tex]W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy[/tex]

 [tex]W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)[/tex]

 [tex]W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0[/tex]

 [tex]W=3420.84*0.25[2401-65536][/tex]

 [tex]W=17084965.5J[/tex]

 [tex]W=17085KJ[/tex]

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Answer:

B energy can't be created or destroyed

Answer:

20 m/s

Explanation:

Applying,

a = (v-u)/t.................... Equation 1

Where a = acceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.

make v the subject of the equation

v = u+at.............. Equation 2

From the question,

Given: u = 0 m/s(start from rest), a = 2 m/s², t = 10 seconds

Substitute these values into equation 2

v = 0+(2×10)

v = 20 m/s

Answer:

[tex] { \huge{s}} = ut + \frac{1}{2} g {t}^{2} \\ 1.5 = 3t + \frac{1}{2} \times 10\times {t}^{2} \\ 1.5 = 3t + 5 {t}^{2} \\ 5 {t}^{2} + 3t - 1.5 = 0 \\ t = 0.3 \: seconds[/tex]

Answer:

magnification is 4

Explanation:

m= image height / object height

m= 12/3

m= 4

Answer:

Capacitance is a derived physical quantity measured in farad

Answer:

"Farad" is another term for the coulomb/volt measurement of capacitance, so both of those options are the correct answer. Amperes and coulomb/second are measurements of electric flow, in other words, how strong a current is.

Answer:

Gaseous; liquid.

Explanation:

A planet can be defined as a large celestial body having sufficient mass to allow for self-gravity and make it assume a nearly circular shape (hydrostatic equilibrium), revolves in an orbit around the Sun in the solar system and has a cleared neighborhood.

Some examples of the planet are Mars, Venus, Earth, Mercury, Neptune, Jupiter, Saturn, Uranus, Pluto, etc.

Basically, the planets are divided into two (2) main categories and these includes;

I. Inner planets: these planets are the closest to the sun and comprises of mercury, venus, earth and mars.

II. Outer planets: these planets are beyond the asteroid belt and comprises of jupiter, saturn, uranus and neptune, from left to right of the solar system.

These outer planets are made mostly of gases (hydrogen and helium) causing them to be less dense than the solid inner planets. These gases are generally known to be less dense in terms of physical properties.

In conclusion, Planets in the solar system can vary from rocky and terrestrial objects to gaseous and liquid object.

However, the set of characteristics which is unique to the outer planets are gaseous and liquid.

Answer:

7. (D) uniformly accelerated vertical motion

8. (A) zero

9. (A) zero

10. (C) parabolic

Answer:

7.Uniformly accelerated vertical motion

8.0m/s²

9.9.8m/s

10.parabolic

11.vertical component.

Answer:

Explanation:

Im going to be using the rules for significant digits properly so I hope you're quite familiar with them. The equation we need for this is

F - f = ma where F is the applied force (our unknown), f is the frictional force, m is the mass, and a is the acceleration. Filling in:

F - 18.2 = 2.25(1.50) and

F = 2.25(1.50) + 18.2  Do the multiplication first and round to get

F = 3.38 + 18.2   The addition rules tell us that we will be rounding to the tenths place after we add to get

F = 21.6 N

[tex]\huge{ \mathrm{  \underline{ Answer }\:  \:  ✓ }}[/tex]

Total force acting on right side = 800 N

Total force acting on left side :

Now, equivalent force acting on the plane is :

And the direction of equivalent force will be the direction of greater force, that is right direction.

Hence, Correct option is :

_____________________________

[tex]\mathrm{ \:TeeNForeveR\:}[/tex]

Answer:

I am not sure if this is the answer

(B) what is the height of the building if it hits the ground after those 5 seconds.

Answer:

Coefficient of kinetic friction = 0.38 (Approx.)

Explanation:

Given:

Applied force = 45 N

Mass of wooden crate = 12 kg

Find:

Coefficient of kinetic friction

Computation:

Coefficient of kinetic friction = Applied force / (Mass)(Acceleration due to gravity)

Coefficient of kinetic friction = 45 / (12)(9.8)

Coefficient of kinetic friction = 45 / 117.6

Coefficient of kinetic friction = 0.3826

Coefficient of kinetic friction = 0.38 (Approx.)

Answer:

B. Current in the circuit is 5.

Ci. Voltage across 7.2 Ω (R₁) is 36 V

Cii. Voltage across 6.2 Ω (R₂) is 31 V

Ciii. Voltage across 8.6 Ω (R₃) is 43 V

Explanation:

We'll begin by calculating the number equivalent resistance in the circuit. This can be obtained as follow:

Resistor 1 (R₁) = 7.2 Ω

Resistor 2 (R₂) = 6.2 Ω

Resistor 3 (R₃) = 8.6 Ω

Equivalent Resistance (R) =?

Since the resistors are in series connection, the equivalent resistance can be obtained as follow:

R = R₁ + R₂ + R₃

R = 7.2 + 6.2 + 8.6

R = 22 Ω

B. Determination of the current.

Voltage (V) = 110 V

Resistance (R) = 22 Ω

Current (I) =?

V = IR

110 = I × 22

Divide both side by 22

I = 110 / 22

I = 5 A

Therefore, the current in the circuit is 5.

Ci. Determination of the voltage across 7.2 Ω (R₁)

Resistor 1 (R₁) = 7.2 Ω

Current (I) = 5 A

Voltage 1 (V₁) =?

V₁ = IR₁

V₁ = 5 × 7.2

V₁ = 36 V

Therefore, the voltage across 7.2 Ω (R₁) is 36 V

Bii. Determination of the voltage across 6.2 Ω (R₂)

Resistor 2 (R₂) = 6.2 Ω

Current (I) = 5 A

Voltage 2 (V₂) =?

V₂ = IR₂

V₂ = 5 × 6.2

V₂ = 31 V

Therefore, the voltage across 6.2 Ω (R₂) is 31 V

Ciii. Determination of the voltage across 8.6 Ω (R₃)

Resistor 3 (R₃) = 8.6 Ω

Current (I) = 5 A

Voltage 3 (V₃) =?

V₃ = IR₃

V₃ = 5 × 8.6

V₃ = 31 V

Therefore, the voltage across 8.6 Ω (R₃) is 43 V

Answer:

C. She can increase the number of wire loops.

Explanation:

The more wire loops the more energy.

For a coil of wire, the magnetic field strength is increased by increasing the number of coils around the nail.

An electromagnet is a soft metal core shaped into a magnet by the passing the electric current through a coil surrounding it.

The end of the wire leading to the S is attached to the positive terminal of a battery. The end of the wire leading to the N is attached to the negative terminal of the battery. The current begins to flow. Current cant be changed to increase magnetic field strength, but the no of coils will definitely increase it.

Thus, To increase the strength of the electromagnet, Lisa can increase the number of wire loops.

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Answer:

I'm pretty sure it is Edwin Powell Hubble and Albert Einstein

Explanation: