a tiger covers a distance of 600 m in 15 minutes what is a speed of a tiger?​

Answer:

B. Is its acceleration constant

Explanation:

Uniform circular motion can be described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. ... An object undergoing uniform circular motion is moving with a constant speed. Nonetheless, it is accelerating due to its change in direction.

Answer:

A gold medal has the (minimum) dimensions of:

diameter = 60mm

thickness = 3mm

So we will work with those dimensions.

The medal is then a cyinder of diameter

D = 60mm = 6cm

and height:

H = 3mm = 0.3cm

Remember that the volume of a cylinder is:

V = pi*(D/2)^2*H

where pi = 3.14

Then the volume of a medal is:

V = 3.14*(6cm/3)^2*0.3cm = 3.768 cm^3

The density of the gold in g/cm^3 is:

d = 19.3 g/cm^3

And remember that:

density = mass/volume

So, if the volume is 3.768 cm^3

Then the mass will be:

mass = density*volume =  19.3 g/cm^3*3.768 cm^3 = 72.7 g

So, a single gold medal would weight 72.7 grams

And each gram of gold costs $40

Then the total cost of the gold medal would be:

value = $40*72.7 = $2,908

Now, if yo think that in the Olympics there are 35 sports (a lot with a large number of players) and near 50 disciplines, they need a lot of gold medals.

And each gold medal costs $2,908

So the total cost (only for the gold medals, ignoring the others) would be to high.

This is why the gold medals are made mostly of silver.

As the spring is stretched, it exerts an upward restoring force f. At maximum extension, Newton's second law gives

∑ F = f - mg = 0   ==>   f = (2.0 kg) (9.8 m/s²) = 19.6 N

By Hooke's law, if k is the spring constant, then

f = kx   ==>   k = f/x = (19.6 N) / (0.15 m) ≈ 130 N/m

A 4.0 kg mass would cause the spring to exert a force of

f = (4.0 kg) (9.8 m/s²) = 39.2 N

which would result in the spring stretching a distance x such that

39.2 N = (130 N/m) x   ==>   x = (39.2 N) / (130 N/m) ≈ 0.30 m ≈ 30 cm

Answer:

a) A = 1.50 m / s²,  B = 1.33 m/s³,  b) a = 12.1667 m / s²,

c)  I = M (1.5 t + 1.333 t²) ,  d)  ΔI = M 2.833   N

Explanation:

In this exercise give the expression for the speed of the rocket

         v (t) = A t + B t²

and the initial conditions

         a = 1.50 m / s² for t = 0 s

         v = 2.00 m / s for t = 1.00 s

a) it is asked to determine the constants.

Let's look for acceleration with its definition

         a = [tex]\frac{dv}{dt}[/tex]

         a = A + 2B t

we apply the first condition t = 0 s

         a = A

         A = 1.50 m / s²

we apply the second condition t = 1.00 s

          v = 1.5 1 + B 1²

          2 = 1.5 + B

          B = 2 / 1.5

          B = 1.33 m/s³

the equation remains

           v = 1.50 t + 1.333 t²

b) the acceleration for t = 4.00 s

           a = 1.50 + 1.333 2t

           a = 1.50 + 2.666 4

           a = 12.1667 m / s²

c) The thrust

           I = ∫ F dt = p_f - p₀

Newton's second law

          F = M a

          F = M (1.5 + 2 1.333 t) dt

we replace and integrate

         I = M ∫ (1.5 + 2.666 t) dt

         I = 1.5 t + 2.666 t²/2

         I = M (1.5 t + 1.333 t²) + cte

in general the initial rockets with velocity v = 0 for t = 0, where we can calculate the constant

         cte = 0

         I = M (1.5 t + 1.333 t²)

d) the initial push

For this we must assume some small time interval, for example between

t = 0 s and t = 1 s

        ΔI = I_f - I₀

        ΔI = M (1.5 1 + 1.333 1²)

        ΔI = M 2.833   N

Answer:

The two skaters move with a speed of 1.73 m/s after the collision in the right direction.

Explanation:

Given that,

The mas of skater 1, m₁ = 80 kg

The speed of skater 1, u₁ = 5 m/s (right)

The mass of skater 2, m₂ = 70 kg

The speed of skater 2, u₂ = -2 m/s (left)

Let v is the magnitude of the two skaters just after they collide. They must have a common speed. So, using the conservation of momentum as follows :

[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\\v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]

Put all the values,

[tex]v=\dfrac{80(5)+70(-2)}{(80+70)}\\\\=1.73m /s[/tex]

So, the two skaters move with a speed of 1.73 m/s after the collision in the right direction.

Answer: The statement that is not an example of convection is (A person gets a suntan on a beach).

Explanation:

There are different modes of heat energy transfer which includes:

--> conduction

--> Radiation and

--> Convection

CONVECTION is a process by which heat energy is transferred in a fluid or air by the actual movement of the heated molecules. The cooler portion of the air surrounding a warmer part exerts a buoyant force on it. As the warmer part of the air moves, it is replaced by cooler air that is subsequently warmed.

Convection in gases is very common and gas expands more than liquid when subjected to high temperature.

--> it is used in bringing about the circulation of fresh air in the room in a process known as ventilation.Here, cool air is constantly being replaced with denser air ( warm air).

-->An electric heater warms a room and Smoke rises above a fire are typical example of convection in gases.

-->Spaghetti is cooked in water: As the water close to the burner warms, it rises to the top and boils. At the same time, cooler water on top moves downward to replace the rising hot water.

--> also the eagle uses convection current to stay afloat in the sky without flapping its wings to conserve energy.

But the option (A person gets a suntan on a beach) is an example of heat transfer through radiation. This is because the sun emits it's rays from the sky down to earth without any material medium unlike others. Therefore, this option is the ODD one out.

Answer:

w = 1,066 rad / s

Explanation:

For this exercise we use Newton's second law

         F = m a

the centripetal acceleration is

         a = w² r

indicate that the force is the mass of the body times the acceleration

        F = m 0.58g = m 0.58 9.8

        F = 5.684 m

we substitute

       5.684 m = m w² r

       w = [tex]\sqrt{5.684/r}[/tex]

To finish the calculation we must suppose a cylinder radius, suppose it has r = 5 m

       w = [tex]\sqrt{ 5.684/5}[/tex]

       w = 1,066 rad / s

Answer:

0.5 V

Explanation:

The electric potential distance between different locations in an electric field area is unaffected by the charge that is transferred between them. It is solely dependent on the distance. Thus, for two electrons pushed together at the same distance into the same field, the electric potential will remain at 1 V. However, the electric potential of one of the two electrons will be half the value of the electric potential for the two electrons.

Answer:

The change in entropy is 1.6 W/K.

Explanation:

Thickness, d = 0.5 cm

Area, A = 1 m^2

T = 25°C

T' = - 10°C

Coefficient of thermal conductivity of glass, K = 0.8 W/mK

The change in entropy is given by

S = Q/T

Here,

[tex]S =\frac{Q}{T}\\\\S = \frac{K A (T - T')}{d(T - T')}\\\\S = \frac{0.8\times 1}{0.5} = 1.6 W/K[/tex]

Answer:

P = 0.14 hp

Explanation:

The power required by the ship is given as:

[tex]P = \frac{Work}{Time} = \frac{Potential\ Eenrgy}{t}\\\\P = \frac{mgh}{t}[/tex]

where,

P = Power = ?

m = mass to pump = (12 lb)(1 kg/2.20 lb) = 5.44 kg

g = acceleration due to gravity = 9.81 m/s²

h = height = 2 m

t = time = 1 s

Therefore,

[tex]P = \frac{(5.44\ kg)(9.81\ m/s^2)(2\ m)}{1\ s}\\\\P = 106.8\ W[/tex]

Converting to horsepower (hp):

[tex]P = (106.8\ W)(\frac{1\ hp}{746\ W})[/tex]

P = 0.14 hp

Answer:

(I)

[tex] { \bf{ {v}^{2} = {u}^{2} - 2as }} \\ {v}^{2} = {0}^{2} - (2 \times 0.5 \times 5) \\ {v}^{2} = 5 \\ { \tt{final \: velocity = 2.24 \: {ms}^{ - 1} }}[/tex]

(ii)

[tex]{ \bf{v = u + at}} \\ 2.24 = 0 + (0.5t) \\ { \tt{time = 4.48 \: seconds}}[/tex]

Answer:

because minerals can be gotten from the bottom

Explanation:

it's self explanatory

Answer:

The impact force is 98000 N.

Explanation:

mass = 10 tons

The impact force is the weight of the object.

Weight =mass x gravity

W = 10 x 1000 x 9.8

W = 98000 N

The impact force is 98000 N.

Answer:

1.5

x 1.0

1.50

x 0.5

075.00

answer: 75.00m

Explanation:

I hope this help

Answer:

ΔE[tex]_{int[/tex],₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J

Explanation:

Given the data in the question;

T[tex]_i[/tex] = 300 K, T[tex]_f[/tex] = 412 K, n = 0.7 mol

since helium is monoatomic;

Cv = (3/2)R, Cp = (5/2)R

W₁ = 0 J [ at constant volume or ΔV = 0]

Now for the first cylinder; from the first law of thermodynamics;

Q₁ = ΔE[tex]_{int[/tex],₁ +  W₁

Q₁ = ΔE[tex]_{int[/tex],₁ = n × Cv × ΔT

we substitute  

Q₁ = ΔE[tex]_{int[/tex],₁ = 0.7 × ( 3/2 )8.314 × ( 412 - 300 )

Q₁ = ΔE[tex]_{int[/tex],₁ = 0.7 × 12.471 × 112

Q₁ = ΔE[tex]_{int[/tex],₁ = 977.7 J

Therefore, ΔE[tex]_{int[/tex],₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J

Answer:

Physical properties are properties that can be measured or observed without changing the chemical nature of the substance. Some examples of physical properties are:

color (intensive)

density (intensive)

volume (extensive)

mass (extensive)

boiling point (intensive): the temperature at which a substance boils

melting point (intensive): the temperature at which a substance melts

Explanation:

Answer:

Option (A) is correct.

Explanation:

A horizontal rope has a length of 5 m and a mass of 0.00145 kg. If a pulse occurs on this string, generating a wavelength of 0.6 m and a frequency of 120 Hz. The tension to which the string is subjected is

mass of string, m = 0.00145 kg

Frequency, f = 120 Hz

wavelength = 0.6 m

Speed = frequency x wavelength

speed = 120 x 0.6 = 72 m/s

Let the tension is T.

Use the formula

[tex]v =\sqrt\frac{T L}{m}\\\\72 = \sqrt\frac{T\times 5}{0.00145}\\\\T = 1.5 N[/tex]

Option (A) is correct.

Answer:

When an object is dropped, the "principal" force that acts on that object is the gravitational force.

Thus, in the absence of air resistance and such, the acceleration of the object will be equal to the gravitational acceleration:

g = 9.8m/s^2

So, when we drop objects in the moon (where there is no air) the acceleration of every object will be exactly the same. (so there is no dependence in the mass or shape of the object)

Thus, if we drop a coin and a feather in the moon, both objects will fall with the same acceleration, and then both objects will hit the ground at the same time.

But if we are in Earth, we can not ignore the air resistance (a force that acts in the opposite direction than the movement of the object)

And this force depends on the shape and mass of the object (for example, something with a really larger surface and really thin, like a sheet of paper will be more affected by this force than a small rock)

Then here, when the air resistance applies, we should expect that the heavier and smaller object (the coin) to be less affected by this force, then the resistance that the coin experiences is smaller, then the coin falls "faster" than the feather.

Answer:

Answer

Correct option is

A

5×10

−6

tesla

I=5A

x=0.2m

Magnetic field at a distance 0.2 m away from the wire.

B=

2πx

μ

0

I

=

2π×0.2

4π×10

−7

×5

=10×5×10

−7

=5×10

−6

tesla

Answer:

a)   W_total = mg (2h + d)   , b)     E_total = - mg (h + d)

Explanation:

a) We must solve this problem in two parts, the first for the accelerated movement and the second for the movement with constant speed

Let's look for work for the part that is in free fall

        y = y₀ + v₀ t - ½ g t²

when he jumps out of a plane his vertical speed is zero

        y =y₀ - ½ g t²

        dy = 0 - ½ g 2t dt

the work in this first part is

        W₁ = ∫ F dy

        W₁ = mg ∫ g t dt

        W₁ = m g² t² / 2

the time it takes to travel the distance y₀-y = h is

         y₀-y = ½ g t²

         t =[tex]\sqrt{2h/g}[/tex]

we substitute

          W₁ = m g² 2h / g

          W₁ = m g 2h

now we look for the work for the part with constant speed

since the velocity is constant let's use the uniform motion ratio

          W₂ = F d

           W₂ = mg d

the total work is

           W_total = W₁ + W₂

           W_total = 2mgh + m gd

           W_total = mg (2h + d)

b) The change in gravitational potential energy

           U = mg Δy

in the part with accelerated movement

           U₁ = mg h

in the part with uniform movement

            U₂ = mg d

the total potential energy is

           E_total = U₁ + U₂

           E_total = - mg (h + d)

Answer:

Angular displacement before it stops = 18 rev

Explanation:

Given:

Speed of fan w(i) = 6 rev/s

Speed of fan (Slow) ∝ = 1 rev/s

Final speed of fan w(f) = 0 rev/s

Find:

Angular displacement before it stops

Computation:

w(f)² = w(i) + 2∝θ

0² = 6² + 2(1)θ

0 = 36 + 2θ

2θ = -36

Angular displacement before it stops = -36 / 2

θ = -18

Angular displacement before it stops = 18 rev

Answer:

Look at explanation

Explanation:

a)Only force acting on the object is gravity, so a=-g (consider up to be positive)

use: v^2=v0^2+2a(y-y0)

plug in givens, at max height v=0

0=400-19.6(H)

Solve for H

H= 20.41m

b) Use: y=y0+v0t+1/2at^2

Plug in givens

0=0+20t-4.9t^2

solve for t

t=4.08 seconds

c) v=v0+at

v=20-39.984= -19.984m/s

Answer: The force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.

Explanation:

Let us assume that

[tex]q_{1} = q_{2} = +q[/tex]

[tex]q_{3} = -q[/tex]

As [tex]q_{3}[/tex] is the negative charge and placed midway between two equal positive charges ([tex]q_{1}[/tex] and [tex]q_{2}[/tex]).

Total distance between [tex]q_{1}[/tex] and [tex]q_{2}[/tex] is 2r. This means that the distance between [tex]q_{1}[/tex] and [tex]q_{3}[/tex], [tex]q_{2}[/tex] and [tex]q_{3}[/tex] = d = r

Now, force action on charge [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.

[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})[/tex]

where,

k = electrostatic constant = [tex]9 \times 10^{9} Nm^{2}/C^{2}[/tex]

Substitute the values into above formula as follows.

[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})\\= 9 \times 10^{9} (\frac{q \times (-q)}{r^{2}})\\= - 9 \times 10^{9} (\frac{q^{2}}{r^{2}})[/tex] ... (1)

Similarly, force acting on [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.

[tex]F_{32} = k \frac{q_{2}q_{3}}{d^{2}}\\= -9 \times 10^{9} \frac{q^{2}}{r^{2}}\\[/tex]   ... (2)

As both the forces represented in equation (1) and (2) are same and equal in magnitude. This means that the net force acting on charge [tex]q_{3}[/tex] is zero.

Thus, we can conclude that the force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.

Answer:

The work done on the block by the spring as it accelerates the block is 4kx².

Explanation:

Let initial distance is x.

It was compressed three times farther and then the block is released, new distance is 3x.

The work done in compressing the spring is given by :

[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)[/tex]

[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\W=\dfrac{1}{2}k((3x)^2-x^2)\\\\W=\dfrac{1}{2}k((9x^2-x^2)\\\\W=\dfrac{1}{2}k\times 8x^2\\\\W=4kx^2[/tex]

So, the work done on the block by the spring as it accelerates the block is 4kx².

Answer:

: No, a clinical thermometer cannot be used to measure the temperature of boiling water because it has a small range and might break due to extreme heat. ... The temperature is around 100 degrees Celsius.

Answer:

Explanation:

Distillation is the separation of multiple LIQUID components based on their different BOILING POINT. As the mixture is heated and the first component SEPARATES, its PURE form travels through the distillation set-up and GOES into a different container

Answer:

a = 6.67 m/s²

Explanation:

F = 10.0 N

m = 1.50 kg

a = ?

F = ma

10.0 = (1.50)a

6.67 = a

First of all, both are not a single sheet of atom. There are many layers of atoms, so the empty part gets beside each other, so there are less empty part. Secondly, there are so many atoms that the probability that they will have empty space at the same place necessary, is negligible.

This was something from logic.

The reason I was taught in my class was that only a limited number of electrons can be in a given orbit, so atoms cannot overlap each other.

Explanation:

Let x = distance of [tex]F_1[/tex] from the fulcrum and let's assume that the counterclockwise direction is positive. In order to attain equilibrium, the net torque [tex]\tau_{net}[/tex] about the fulcrum is zero:

[tex]\tau_{net} = -F_1x + F_2d_2 = 0[/tex]

[tex] -m_1gx + m_2gd_2 = 0[/tex]

[tex]m_1x = m_2d_2[/tex]

Solving for x,

[tex]x = \dfrac{m_2}{m_1}d_2[/tex]

[tex]\:\:\:\:=\left(\dfrac{105.7\:\text{g}}{65.7\:\text{g}} \right)(13.8\:\text{cm}) = 22.2\:\text{cm}[/tex]