A ball is projected horizontally from the top of a bertical building 25.0m above the ground level with an initial velocity of 8.25m/s . Calculate :
a. How long is the ball in flight before striking the ground
b. How far from the building does the ball strike the ground

Answer:

B illustrates refraction

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

Answer:

a) Pascal is a derived unit because it is derived from the unit of force and area

b)Mass is a fundamental quantity because it doesn't depends upon others physical quantity and made up of only one unit

c) unit of power is a derived unit because they are dependent quantities

D) unit of length is a fundamental unit because it cannot be expressed in terms of another quantity.

Answer:

c

Explanation:

Answer:

This question is misleading since

1/C = 1/C1 + 1/C2 + 1/C3 + 1/C4

so it would have to be less than the smallest individual capacitance

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:

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:

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:

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

Explanation:

it just it

Answer:

third one

Explanation:

i cant think of anymore

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:

Increasing

Explanation:

I Hope it Helps

Answer:

cant see picture

Explanation:

Answer:

please add picture so i can help you

Explanation:

Answer:

Explanation:

a)The average acceleration of the sled will be 6.198 m/sec

b)The amount of time required to travel at the highway's 100 km/h speed limit will be 4.48 sec.

c)The distances covered when the vehicle hits its top speed of 270 km/h will be 453.77 m.

The rate of velocity change concerning time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.

The pace at which a body's velocity varies is represented by acceleration, which is a vector quantity.

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

The given data in the problem is given by ;

u is the initial speed of the rocket=  0 km/h

v is the final speed of the rocket=  270 km/h

t is the time interval= 12.1 seconds

a is the average acceleration of the ball=? m/sec²

to is the time it takes to reach the speed of 100 km/h

s is the distance traveled when it reaches the final speed

a)

The average acceleration of the sled is found as;

a = (v-u)/t

a=(75-0)/12.1

a = 6.198 m/sec

b)

The amount of time required to travel at the highway's 100 km/h speed limit;

From Newton's second equation of motion;

v =u+at

27.7 = 0 + 6.198t

t = 27.7 / 6.198

t = 4.48 sec

c)

The distances covered when the vehicle hits its top speed of 270 km/h;

v²=u²+2as

75²=0+2(6.198)s

s=(75)²/2(6.198)

s=453.77 m

Hence the average acceleration of the sled, the amount of time required to travel at the highway's 100 km/h speed limit, and the amounts of time required to travel at the highway's 100 km/h speed limit will be  6.198 m/sec,4.48 sec and 453.77 m respectively.

To learn more about acceleration, refer to the link;

https://brainly.com/question/2437624

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Answer:

 ΔR = 5 s

Explanation:

The absolute uncertainty or error in an expression is

       ΔR = | [tex]\frac{dR}{dB}[/tex] | ΔB + | [tex]\frac{dR}{dA}[/tex] | ΔA

the absolute value guarantees to take the unfavorable case, that is, the maximum error.

We look for the derivatives

       [tex]\frac{dR}{dB}[/tex] = 1

       [tex]\frac{dR}{dA}[/tex] = -1

we substitute

       ΔR = 1 ΔB + 1 ΔA

of the data

       ΔB = 3 s

       ΔA = 2 s

       ΔR = 3 + 2

       ΔR = 5 s

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

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:

True

Explanation:

The tensile stress, σ, on a solid cylindrical wire is given by the following relationship;

[tex]\sigma = \dfrac{F_t}{A_o}[/tex]

Where;

[tex]F_t[/tex] = The tensile force

[tex]A_o[/tex] = The original cross sectional area of the cylindrical wire = π·R²

R = The radius of the wire

Therefore;

[tex]F_t[/tex] = σ × [tex]A_o[/tex] = σ × π × R²

Therefore, the tensile force is directly proportional to the square of the radius of the cylindrical wire, and as the radius of the wire increases, which is by increasing the thickness of the wire, the tensile force is largely increased

The correct option is; True.

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

f=G m1m2/r2

Answer:

Answer:

a)    I = 13.38 kg m / s, b)    F = 1,373 10³ N

Explanation:

The impulse is given by the relation

          I = ∫ F dt = Δp

          I = p_f -p₀

          I = m (v_f - v₀)

take the ball's exit direction as positive, whereby the ball velocities

v₀ = -90mph, the final velocity v_f = + 54 m / s

Let's reduce the units to

         I = 0.142 [54- (-40.23) ]

      the SI system

        v₀ = - 90 mph (1609.34 m / 1 mile) (1h / 3600 s = -40.23 m / s

        m = 142 g (1kg / 1000) = 0.142 kg

we calculate  

          I = 0.142 [54- (-40) ]

          I = 13.38 kg m / s

b) let's use the definition of momentum

         I = ∫ F .dt

         I = F ∫ dt

         F = I / t

         F = 13.38 / 0.008

         F = 1,373 10³ N

Answer:

Frequency of the tuning fork[second resonance] = 512 Hz (Approx.)

Explanation:

Given:

Speed of sound = 340 m/s

Length of resonance position above a closed air column = 49.8 cm = 0.498 m

Find:

Frequency of the tuning fork

Computation:

Frequency of the tuning fork[second resonance] = 3v / 4l

Frequency of the tuning fork[second resonance] = 3(340) / 4(0.498)

Frequency of the tuning fork[second resonance] = 512.04

Frequency of the tuning fork[second resonance] = 512 Hz (Approx.)

Answer:

The frequency is 512 Hz.

Explanation:

speed, v = 340 m/s

length, L = 49.8 cm = 0.498 m

let the frequency is f.

[tex]f =\frac{3 v}{4 L}\\\\f = \frac{3 \times 340 }{4\times 0.498}\\\\f = 512 Hz[/tex]

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

Answer:

I think it's option D

Explanation:

I think it's option D but not so sure

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

Answer:

The position is 8.18cm from the mirror.

Nature is b=virtual

Size is 1.82cm

Explanation:

Note that for a convex mirror, the image distance and the focal length are negative;

Given

Object height H0 = 4cm

object distance u = 18cm

Radius of curvature R = 30cm

Since f = R/2

f = 30/2

f = -15cm

Recall that:

[tex]\frac{1}{f} =\frac{1}{u}+ \frac{1}{v}\\\frac{1}{-15}=\frac{1}{18}+\frac{1}{v} \\\frac{1}{v} =\frac{1}{-15} -\frac{1}{18}\\ \frac{1}{v} = \frac{-18-15}{270}\\\frac{1}{v} = \frac{-33}{270}\\v=\frac{-270}{33}\\v=-8.18cm[/tex]

Since the image distance is negative, this shows that the image is a virtual image.

To get the size:

[tex]\frac{H_1}{H_0}=\frac{v}{u}\\\frac{H_1}{4}=\frac{8.18}{18}\\18H_i=32.72\\H_i=\frac{32.72}{18}\\H_i= 1.82cm[/tex]