What does E=mc2 stand for?

Answer:

the answer is nearly 5.655 [tex]ms^{-2}[/tex]

Explanation:

Given,

[tex]v_{x}=2.7 ms^{-1}[/tex]

[tex]v_{y}=13.3 ms^{-1}[/tex]

[tex]t=2.4 s[/tex]

[tex]a_{x}=\frac{2.7}{2.4}=1.125 ms^{-2}[/tex] (as  [tex]a=\frac{v-u}{t}[/tex])

[tex]a_{y}=\frac{13.3}{2.4}=5.542 ms^{-2}[/tex]

[tex]a=\sqrt{a_{x}^{2}+a_{y}^{2} }[/tex]

[tex]=\sqrt{1.125^{2}+5.542^{2} }[/tex]

[tex]=5.655 ms^{-2}[/tex]

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Answer:

6.94 km/hr

Explanation:

m/s to km/hr -> Multiply by 18/5

25/(18/5)

=> 25 x 5/18

=> 125/18 km/hr

=> 6.94 km/hr

Answer: 90,000 m = 90 km

Explanation:

Given information

Time = 1 hour

Speed = 25 m/s

Given expression deducted from the given information

Distance = speed × time

Convert units of time

1 hour = 60 minutes

1 minute = 60 seconds

1 hour = 60 × 60 = 3600 seconds

Substitute values into the expression

Distance = 25 × 3600

Simplify by multiplication

Distance = [tex]\boxed{90,000 m=90km}[/tex]

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Answer:

q.1 : Air near candle gets heated up and after this it rises by convection so the thermometer B will receive more heat than the thermometer A So, according to the given condition thermometer B will show a greater rise in temperature.

q.2 : x is the pure sample of compound . y is the pure sample of element . z is the mixture of different elements

q.3 : the saliva contains an enzyme salivary amylase (ptyalin) which converts starch in roti into maltose, isomaltose and small dextrins called a-dextrin.

Answer:

B 5s

Explanation:

Because of the Displacement in the nth second of the free fall is 

Snth=21g(t12−t22)

Given that (tn−tn−1)=1

Displacement in 3 seconds of the free fall 

S=21gt2

S=21×10×32

S=45m

Given that: Snth=45

On solving that we get:

t1=5sec

Answer:

Answer is in the picture.

Explanation:

Answer is in the picture.

Answer:

[tex]\theta=34 \textdegree[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=55kg[/tex]

Angle [tex]\theta =28.0[/tex]

Coefficient of static friction [tex]\alpha =0.680[/tex]

Generally, the equation for Newtons second Law is mathematically given by

For

[tex]\sum_y=0[/tex]

[tex]N=mgcos \theta[/tex]

for

[tex]\sum_x=0[/tex]

[tex]F_{s}=mgsin\theta[/tex]

Where

[tex]F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta[/tex]

[tex]F_{s}=0.68*55*9.8*cos 28[/tex]

[tex]F_{s}=323.62N[/tex]

Therefore

[tex]\alpha mgcos \theta=mg sin \theta[/tex]

[tex]\theta=tan^{-1}(0.68)[/tex]

[tex]\theta=34 \textdegree[/tex]

(a) The static frictional force which holds the box in place is 323.62 N.

(b) The maximum angle that the incline can make with the horizontal is 34.2⁰.

The net force applied to keep the box at rest must be zero in order for the box to remain in equilibrium position. Apply Newton's second law of motion to determine the net force.

∑F = 0

The static frictional force is calculated as follows;

Fs = μFncosθ

Fs = 0.68 x (55 x 9.8) x cos28

Fs = 323.62 N

Fn(sinθ) - μFn(cosθ) = 0

mg(sinθ) - μmg(cosθ) = 0

μmg(cosθ) = mg(sinθ)

μ(cosθ) = (sinθ)

μ = sinθ/cosθ

μ = tanθ

θ = tan⁻¹(μ)

θ = tan⁻¹(0.68)

θ = 34.2⁰

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Answer:

The answer is 312.5j

Explanation:

The kinetic energy (KE):

KE=1/2*m*v^2

M= mass of the object

v= velocity of the object

We have;

m=25g

v=5m/s

KE=1/2*25g*5^2m/s

KE =312.5j

The acceleration of the block over the rough surface is 1.22625 m/s²

The process through which the acceleration is obtained is presented as follows of approach to

The given parameters are;

Mass of block, m = 2 kg

Nature of the surface of the quarter circle = Frictionless

The length of the horizontal, d = 8 m

The coefficient of friction of the horizontal surface, μ = 0.4

The unknown parameter;

The acceleration of the block over the rough surface

Method;

Find the work done by friction to stop the block and divide the result by the mass of the block

The work done by friction, [tex]W_f[/tex] = (Force of friction) × (Distance the block moves on the rough surface before coming to rest)

[tex]\mathbf{W_f}[/tex] = [tex]\mathbf{F_f}[/tex] × d

[tex]F_f[/tex] = Normal reaction of surface on block, [tex]N_r[/tex] × μ

Normal reaction on block, [tex]\mathbf{N_r}[/tex] = Weight of block

[tex]\mathbf{N_r}[/tex] ≈ 2 kg × 9.81 m/s² = 19.62 N

Therefore;

The work done by friction [tex]\mathbf{W_f}[/tex] = [tex]\mathbf{F_f}[/tex] × d = [tex]\mathbf{N_r}[/tex] × μ × d

[tex]\mathbf{W_f}[/tex] = 19.62 N × 0.4 × 8 m = 62.784 J

The work done by the block, W = Force, F × d

Force, F = m × a

Where;

a = The acceleration of the block

According to the principle of conservation of energy, we have;

[tex]\mathbf{W_f}[/tex]  = W

∴ 19.62 J = 2 kg × a × 8 m

a = 19.62/(2 kg × 8 m) = 1.22625 m/s²

The acceleration of the block over the rough surface, a = 1.22625 m/s²

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Answer:

Subrahmanyan Chandrasekhar

Answer:

D. Subrahmanyan Chandrasekhar

Explanation:

[tex]19.2\:\text{m/s}[/tex]

Explanation:

At the top of the tree, the velocity of the pebble is purely horizontal so we can calculate it as

[tex]v_{y} = v_{0y} = v_0\cos 40° = (25\:\text{m/s})(0.766)[/tex]

[tex]\:\:\:\:\:= 19.2\:\text{m/s}[/tex]

The height of the tree is approximately 12.5 meters when velocity of the pebble as it passes over the top of the tree.​

Let's calculate the height of the tree step by step:

Given:

Initial velocity (v0) = 25 m/s

Launch angle (θ) = 40° above the horizontal

Time after launch (t) = 2 seconds

Acceleration due to gravity (g) = -9.8 m/s² (negative because it acts downward)

Step 1: Calculate the vertical component of the initial velocity (Vy):

Vy = v0 * sin(θ)

Vy = 25 m/s * sin(40°)

Vy ≈ 25 m/s * 0.6428 ≈ 16.07 m/s (rounded off to two decimal places)

Step 2: Calculate the vertical displacement (change in height) of the pebble after 2 seconds:

d = vot + (1/2)at²

d = (16.07 m/s) * (2 s) + (1/2) * (-9.8 m/s²) * (2 s)²

d ≈ 32.14 m - 19.6 m

d ≈ 12.54 m (rounded off to two decimal places)

Step 3: The height of the tree is equal to the vertical displacement of the pebble:

Height of the tree ≈ 12.54 m ≈ 12.5 m (rounded off to one decimal place)

The height of the tree is approximately 12.5 meters.

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Answer:

ans is c

Explanation:

chk photo

The axial magnetic field 15 cm from the center of the loop is approximately 21 μT. Hence, option (c) is correct.

In the vicinity of a magnet, an electric current, or a shifting electric field, there is a vector field called a magnetic field where magnetic forces can be seen. Electric charges in motion and the intrinsic magnetic moments of elementary particles connected to the fundamental quantum characteristic known as spin create a magnetic field.

Given parameters:

Radius of the circular loop: r = 10 cm = 0.10 m.

Current passing through the loop: I = 20 A.

Axial distance of the point: z = 15 cm = 0.15 m.

Hence, the axial magnetic field at that point:

B = (μ₀/4π) 2πR²I/(z²+R²)^(3/2)

By putting these values, we get B = 21 × 10⁻⁶ T = 21 μT.

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Answer:

Hi I hope this is correct!

Explanation:

To find average velocity you can use the formula av = (v1 + v2) / 2

*I converted everything into m/s because that it usually the measurement for velocity*

v1 = initial velocity = 8.49376 m/s , v2 = final velocity = 33.528 m/s

av = 8.49376 + 33.528 / 2

    = 21.01088 m/s

*If you were required to leave the final answer in mph here it is

av = 19 + 75 / 2

    = 47 mph

Hope this helps! Best of luck <3

Explanation:

hope it helps you

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Answer:

SI is used in most places around the world, so our use of it allows scientists from disparate regions to use a single standard in communicating scientific data without vocabulary confusion

MKS stands for metre , kilogram and second

Answer:

buriret i belive

Explanation:

.

Answer:

The blue solution is named copper sulfate

Answer:

don't know what class are you you are using which mobile or laptop

Answer:

The first is the electric field, which describes the force acting on a stationary charge and gives the component of the force that is independent of motion. The magnetic field, in contrast, describes the component of the force that is proportional to both the speed and direction of charged particles.

We have that the total distance covered by the runner is

[tex]d_t=20.723miles[/tex]

The total distance covered by the runner is a sum of all miles covered by the runner

Therefore

With

[tex]d_t[/tex]=Total distance

[tex]d_t=d_1+d_2+d_3\\\\d_t=9.5+8.89+2.333[/tex]

[tex]d_t=20.723miles[/tex]

in conclusion

The total distance covered by the runner is

[tex]d_t=20.723miles[/tex]

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Answer:

Their is a direct relationship between the number of batteries and the increase in power. The voltage is the product of the number of batteries and the voltage which is 9 volts. As the batteries touch ends the voltages of all three combines.

Explanation:

Answer:

26 km

Explanation:

Let's say our "cable" has a cross section of 1 m²

Then each meter of cable would weight 7900(9.8) = 77420 N

A Pascal is a Newton per square meter

2 x 10⁹ / 77420 = 25840 m or about 26 km or about 16 miles

Answer:

option A

Explanation:

simple harmonic motion

Answer:

random motion I think not sure

Blood comes into the right atrium from the body, moves into the right ventricle and is pushed into the pulmonary arteries in the lungs. After picking up oxygen, the blood travels back to the heart through the pulmonary veins into the left atrium, to the left ventricle and out to the body's tissues through the aorta.

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have a nice day

 wavefront is the long edge that moves, for example, the crest or the trough

Answer:

1km=1000m

700km=

700×1000=700000

=700000metres

hope this helps

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let T represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, [tex]F_{f2}[/tex] = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = [tex]F_{f2}[/tex] × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

[tex]T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27[/tex]

[tex]Frictional \ force, F_{f2} = \left (981 - \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times 0.25 \right) \approx 219.92[/tex]

The frictional force on the block W₂, [tex]F_{f2}[/tex] ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ [tex]F_{w2}[/tex] = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + [tex]\mathbf{F_{w2}}[/tex]

The frictional force from the ground, [tex]\mathbf{F_{f1}}[/tex] = N×μ + [tex]\mathbf{F_{f2}}[/tex] = P

Where;

P = The horizontal force applied to the block

P = (W₁ + [tex]\mathbf{F_{w2}}[/tex]) × μ + [tex]\mathbf{F_{f2}}[/tex]

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

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The generator is 124.6 volts RMS.

Answer:

B. Increases.

Explanation:

[tex]{ \bf{F = \frac{m {v}^{2} }{r} }} \\ { \bf{F \: \alpha \: {v}^{2} }}[/tex]

Keeping mass, and radius constant, speed or velocity is directly proportioanal to centripetal force.

,In order to increase the speed of an object on a circular path, YOU have to increase the centripetal force acting on it.