An object has a kinetic energy of 8.8J and a mass of 450g. How fast is the object moving?

Answer:

B illustrates refraction

B. Watts

Then j/s is the rate of transferring energy or doing work. Its unit is the Watt, equivalent to 1 Joule per second.

Answer:

see the explanation below

Explanation:

Momentum is a product of the mass of a particle and its velocity.

and also, momentum is a vector quantity; i.e. it has both magnitude and direction.

Now a plane in the air has both magnitude and velocity

When the plane lands the velocity will amount to zero although the mass is still very much intact

Now the mass* zero velocity= zero

Hence when a plane lands the momentum is zero

Answer:

The diagram assigned B

explanation:

Check the direction of the two vectors, their resultant must be in the same direction.

Answer:

Explanation:

We need to find the x-components of each of these vectors and then add them together, then we need to find the y-components of these vectors and then add them together. Let's get to that point first. That's hard enough for step 1, dontcha think?

The x-components are found by multiplying the magnitude of the vectors by the cosine of their respective angles, while the y components are found by multiplying the magnitude of the vectors by the sine of their respective angles.

Let's do the x-components for all the vectors first, so we get the x-component of the resultant vector:

[tex]F_{1x}=12 cos0[/tex] and

[tex]F_{1x}=12[/tex]

[tex]F_{2x}=9cos90[/tex] and

[tex]F_{2x}=0[/tex]

[tex]F_{3x}=15 cos126.87[/tex] and

[tex]F_{3x}=-9.0[/tex]  (the angle of 126.87 is found by subtracting the 53.13 from 180, since angles are to be measured from the positive axis in a counterclockwise fashion).

That means that the x-component of the resultant vector, R, is 3.0

Now for the y-components:

[tex]F_{1y}=12sin0[/tex] and

[tex]F_{1y}=0[/tex]

[tex]F_{2y}=9sin90[/tex] and

[tex]F_{2y}=9[/tex]

[tex]F_{3y}=15sin126.87[/tex] and

[tex]F_{3y}=12[/tex]

That means that the y-component of the resultant vector, R, is 21.

Put them together in this way to find the resultant magnitude:

[tex]R_{mag}=\sqrt{(3.0)^2+(21)^2}[/tex] which gives us

[tex]R_{mag}=21[/tex] and now for the angle. Since both the x and y components of the resultant vector are positive, our angle will be where the x and y values are both positive in the x/y coordinate plane, which is Q1.

The angle, then:

[tex]tan^{-1}(\frac{21}{3.0})=82[/tex] degrees, and since we are QI, we do not add anything to this angle to maintain its accuracy.

To sum up: The resultant vector has a magnitude of 21 N at 82°

Answer:

a. X: Slower in gases than liquids Y: Faster in solids than gases Z: Velocity depends on medium.

Explanation:

Speed of sound is fastest in solids. Sound waves travel more quickly in solid, than of liquid and gases. Sound waves travel most slowest in gases. Speed of sound varies significantly and it depends upon medium it is travelling through.  In more rigid medium sounds velocity will be faster.

Answer:

muddy water is a heterogeneous mixture, which is Suspension.

Answer:

average speed is 60km/h

Explanation:

you sum up the speed attained in each distance covered and divide it by 2 to get your answer

Answer:

[tex] Energy, \; E = 2.6504 * 10^{-34} \; Joules [/tex]

Explanation:

Given the following data;

Frequency = 4.0 x 10⁹ Hz

Planck's constant, h = 6.626 x 10-34 J·s.

To find the energy of the electromagnetic wave;

Mathematically, the energy of an electromagnetic wave is given by the formula;

E = hf

Where;

E is the energy possessed by a wave.

h represents Planck's constant.

f is the frequency of a wave.

Substituting the values into the formula, we have;

[tex] Energy, \; E = 4.0 x 10^{9} * 6.626 x 10^{-34} [/tex]

[tex] Energy, \; E = 2.6504 * 10^{-34} \; Joules [/tex]

Answer:  

GIVEN:

v₀=0ms⁻¹

a= 8.1ms⁻²

t= 19.4s

REQUIRE:

d=?

CALCULATUION:

as we know,

d=v₀t+1/2at²

by putting values

d=0ms⁻¹×19.4s+1/2×8.1ms⁻²×(19.4s)²

d=0m+1/2×8.1ms⁻²×376.36s²

d=1/2×3048.516m

d=1524.258m

d≈1524m

Answer:

i. Work done by the gas as it expands is approximately 1,900 J

ii. The total heat supplied is approximately 4, 576 J

iii. The change in internal energy is approximately 2,772 J

Explanation:

The constant pressure of the helium gas, P = 1.0 × 10⁵ Pa

The initial and final pressure of the gas, T₁, and T₂ = 2°C (275.15 K) and 112°C (385.15 K) respectively

The number of moles of helium in the sample of helium gas, n = 2 moles

The volume occupied by the gas at state 1, V₁ = 45 L

i. By ideal gas law, we have;

P·V = n·R·T

Therefore;

[tex]V = \dfrac{n \cdot R \cdot T}{P}[/tex]

Plugging in the values gives;

[tex]V_2 = \dfrac{n \cdot R \cdot T_2}{P}[/tex]

Where;

V₂ = The volume of the gas at state 2

Therefore;

[tex]V_2 = \dfrac{2 \cdot 8.314 \cdot 385.15}{1.0 \times 10^5} \approx 0.064[/tex]

The volume of the gas at state 2, V₂ ≈ 0.064 m³ = 64 Liters

Work done by the gas as it expands, W = P × (V₂ - V₁)

∴ W ≈ 1.0 × 10⁵ Pa × (64 L - 45 L) = 1,900 J

Work done by the gas as it expands, W ≈ 1,900 J

ii. The total heat supplied, Q = Cp·n·ΔT

∴ Q = 20.8 J/(mol·K) × 2 moles × (385.15 K - 275.15 K) = 4,576 J

The total heat supplied, Q = 4, 576 J

iii. The change in internal energy, ΔU = Cv·n·ΔT

∴ ΔU = 12.6 J/(mol·K) × 2 moles × (385.15 K - 275.15 K) = 2,772 J

The change in internal energy, ΔU = 2,772 J

Answer:

77.625 cm

Explanation:

The given distance of the image behind the convex mirror, v = 11.5 cm

The focal length of the mirror, f = 13.5 cm

The mirror formula for convex mirror is given as follows;

[tex]\dfrac{1}{u} - \dfrac{1}{v} = -\dfrac{1}{f}[/tex]

Where;

u = The distance from the statue to the mirror

Therefore, we get;

[tex]\dfrac{1}{u} = -\dfrac{1}{f} + \dfrac{1}{v}[/tex]

Plugging in the values gives;

[tex]\dfrac{1}{u} = -\dfrac{1}{13.5} + \dfrac{1}{11.5} = \dfrac{8}{621}[/tex]

∴ The distance from the statue to the mirror, u = 621/8 cm = 77.625 cm.

Answer:

ΔP = 1875 Pa,   P₂ = P₁ - 1875

Explanation:

Let's use Bernoulli's equation, with the subscript 1 for the widest Mars and the subscript 2 for the narrowest part, suppose that the pipe is horizontal

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

          P₁ -P₂ = ½ ρ (v₂² - v₁²)

suppose the fluid is water

          P₁ - P₂ = ½ 1000 (2² - 0.5²)

         ΔP = 1875 Pa

this is the pressure difference between the two sections

the pressure in the narrowest section is

           P₂ = P₁ - 1875

Answer:

It is simile uses "like" or "as": metaphor does not

Explanation:

it just it

Answer:

third one

Explanation:

i cant think of anymore

KE = (0.5) m v²

given that : v = speed of the bottle in each case = 4 m/s when m = 0.125 kg

KE = (0.5) m v² = (0.5) (0.125) (4)² = 1 J

when m = 0.250 kg KE = (0.5) m v² = (0.5) (0.250) (4)² = 2 J

when m = 0.375 kg KE = (0.5) m v² = (0.5) (0.375) (4)² = 3 J

when m = 0.0.500 kg KE = (0.5) m v² = (0.5) (0.500) (4)² = 4 J

The gravitational force between 2 bodys decreases with distance between the two bodies.

f=G m1m2/r2

Answer:

Answer:

Mass, M = 4.859 Kg

Explanation:

Given the following data;

Radius, r = 0.225 m

Moment of inertia, I = 0.123 kgm²

To find the mass;

Mathematically, the moment of inertia is given by the formula;

I = ⅖Mr²

Making M the subject of formula, we have;

Cross-multiplying, we have;

2I = Mr²

[tex] M = \frac {2I}{r^{2}} [/tex]

Substituting into the formula, we have;

[tex] M = \frac {2*0.123}{0.225^{2}} [/tex]

[tex] M = \frac {0.246}{0.050625} [/tex]

Mass, M = 4.859 Kg

Answer:

Distance = 450 kilometres

Explanation:

Given the following data;

Speed = 90 km/h

Time = 5 hours

To find the distance covered by the train;

Mathematically, the speed of an object or body is given by the formula;

[tex] Speed = \frac {distance}{time} [/tex]

Making distance the subject of formula, we have;

Distance = speed * time

Substituting into the formula, we have;

Distance = 90 * 5

Distance = 450 kilometres

The solubility of the sample will decrease

Answer:

work done = mgh

350×10×2

7000J

The work done by the forklift truck as it lifts the box to the given height is 6860J.

Work is simply referred to as the displacement of an object when a push or pull force is applied to the object. It is the energy transferred from or to an object when force is applied to it along a displacement.

It is expressed as;

W = F × s

Where F is force and s is displacement

Given the data in the question;

We substitute our given values into the expression above.

W = F × s

But F = Weight = mass × acceleration due to gravity = mg

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

Hence,

W = mg × s

W = 350kg × 9.8m/s² × 2m

W = 6860kgm²/s²

W =  6860J

Therefore, the work done by the forklift truck as it lifts the box to the given height is 6860J.

Learn more about Work: https://brainly.com/question/9942439

At the ball's highest point, it has no vertical velocity, so the 6 m/s is purely horizontal. A projectile's horizontal velocity does not change, which means the ball was initially thrown with speed v such that

v cos(53°) = 6 m/s   ==>   v = (6 m/s) sec(53°) ≈ 9.97 m/s

The player shoots the ball from a height of 2.0 m, so that the ball's horizontal and vertical positions, respectively x and y, at time t are

x = (9.97 m/s) cos(53°) t = (6 m/s) t

y = 2.0 m + (9.97 m/s) sin(53°) t - 1/2 gt ²

Find the times t for which the ball reaches a height of 3.00 m:

3.00 m = 2.0 m + (9.97 m/s) sin(53°) t - 1/2 gt ²

==>   t ≈ 0.137 s   or   t ≈ 1.49 s

The second time is the one we care about, because it's the one for which the ball would be falling into the basket.

Now find the distance x traveled by the ball after this time:

x = (6 m/s) (1.49 s) ≈ 8.93 m

Answer:

I think it's option D

Explanation:

I think it's option D but not so sure

Answer:

This may refer to a situation like:

"one person pushes a box, if there is equal and opposite reaction why the box moves and the person does not?"

Remember the second Newton's law:

F = m*a

suppose that the mass of the person is 3 times the mass of the box.

So, if the box has a mass M, the person will have a mass 3*M

Then the Newton's equation for the box when the person pushes with a force F is:

F = M*a

solving for the acceleration, we get:

F/M = a

While the person is also pushed by the box with a force with the same magnitude, then the equation for the person is:

F = (3*M)*a'

Solving for the acceleration, we get:

F/(3M) = a'

Now we can compare the acceleration of the box (F/M) with the acceleration of the person (F/3M).

Is easy to see that the acceleration of the box is 3 times the acceleration of the person.

So regardless of the fact that both the box and the person experience a force with the same magnitude, the box will move more due to this force.

This is why in situations like this, the forces do not seem balanced.

Answer:

Explanation:

Acceleration is equal to the change in velocity over the change in time, or

[tex]a=\frac{v_f-v_i}{t}[/tex] where the change in velocity is final velocity minus initial velocity. Filling in:

[tex]3.0=\frac{v_f-(-3.0)}{6.0}[/tex] Note that I made the backward velocity negative so the forward velocity in our answer will be positive.

Simplifying that gives us:

[tex]3.0=\frac{v_f+3.0}{6.0}[/tex] and then isolating the final velocity, our unknown:

3.0(6.0) = v + 3.0 and

3.0(6.0) - 3.0 = v and

18 - 3.0 = v so

15 m/s = v and because this answer is positive, that means that the car is no longer rolling backwards (which was negative) but is now moving forward.

Answer:

reduce the velocity of collision

"Rotating wheel" is meant to indcate that the wheel is already rotating at the start. Denote the initial angular velocity by ω₀. Then the angular displacement θ at time t is

θ = ω₀t + 1/2 αt ²

while the angular velocity ω is

ω = ω₀ + αt

It takes 5 s for the wheel to rotate 38 times, or turn a total of (2π rad/rev) (38 rev) = 76π rad, as well as to reach an angular velocity of 79 rad/s, so that

76π rad = ω₀ (5 s) + 1/2 α (5 s)²

79 rad/s = ω₀ + α (5 s)

Solve the second equation for ω₀ and substitute into the first equation, then solve for α :

ω₀ = 79 rad/s - α (5 s)

==>   76π rad = (79 rad/s - α (5 s)) (5 s) + 1/2 α (5 s)²

==>   76π rad = (79 rad/s) (5 s) - 1/2 α (5 s)²

==>   1/2 α (5 s)² = (79 rad/s) (5 s) - 76π rad

==>   α = ((79 rad/s) (5 s) - 76π rad) / (1/2 (5 s)²)

==>   α ≈ 12.5 rad/s²

Answer:

p = -9 m

Explanation:

For this exercise we use the equation of the geometric optics constructor

         [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

where f is the focal length, p and q are the distance to the object and the image, respectively.

The mirrors the focal length is

         f = R / 2

         f = 2/2

         f = 1 m

the magnification is

         m =[tex]\frac{h'}{h} = - \frac{q}{p}[/tex]

indicates that the image was reduced h ’= h/10 implies that m = 1/10  

          [tex]\frac{1}{10} = - \frac{q}{p}[/tex]

we write our system of equations

            p = -10q

            1/1 = [tex]\frac{1}{p} + \frac{1}{q}[/tex]

we substitute

           1 = [tex]\frac{1}{p} - \frac{10}{p}[/tex]

           1 = 1/p  (1 - 10)

           1 = -9 / p

           p = -9 m