how you brother and sister​

Answer:

[tex]a_{x} = 0.342 \frac{m}{ {s}^{2} } \: "x \: component \: of \: acceleration"

\: \\ a_{y} = 1.02 \: \frac{m}{ {s}^{2} } \: "y \: component \: of \: acceleration" [/tex]

Explanation:

Given a mass of 63.9kg with a total applied force of 125 N at a 31.5° inclination relative to the horizontal and vertical, and a horizontal resistance force of 84.8 N. The components of acceleration can be calculated as follows:

Force = mass × acceleration

Vertical component of a vector = vector × Sin(angle)

Horizontal component of a vector = vector × Cos(angle)

extra:

V = √x^2+y^2

θ (angle, "theta") = arctan or inverse tan(y/x) (For this instance. Theta is usually an angle measure though)

x = V cos θ

y = V sin θ

____________________

To find the vertical component of acceleration we must first take the vector quantity of our force which is 125 N (Newtons) and it's angle of 31.5°.

Vertical force = 125 × sin(31.5 degrees) = 125 × 0.5225 = 65.312 N [aka the normal force]

This works because force is a vector, it has both direction(by the angle of force), and magnitude(it's quantity).

Then take the force equation, and rearrange it to solve for the upward acceleration: F = m × a → a = F/m.

a = 65.312/63.9 ≈ 1.02 m/s^2.

Then for the horizontal acceleration, take the cosine for the horizontal part

Answer:muscle strength

Explanation: bc well you would need it

Answer:

it gets colder

Explanation:

because if you put anything on a table top it will become colder

Answer:

114 m/s

Explanation:

see the image below

Answer:

  9.877 m/s^2

Explanation:

The acceleration can be computed from ...

  d = (1/2)at^2

  (1600 m) = (1/2)a(18 s)^2

  a = (1600/162) m/s^2 ≈ 9.877 m/s^2

Answer:

54mph

Explanation:

Answer:

Acceleration = -12/17 m/s ^2

Explanation:

Name the variables

V = final velocity( velocity after the train stops) u= intial velocity(velocity when the train was moving at constant speed, before brakes are applied)

A = acceleration( in this case, it is deceleration)

T = time ( time taken to fully stop)

Second Step: What equation should we use?

The 3 Big Motion in One Dimension Formulas are:

V = u+at

v^2 = u^2 + 2as

S = ut+ (1/2) at^2

For this question, we will be using the first one.

Third Step: Solve

V = 0 (the final velocity is 0, because the train stopped)

U = 60 (the train was going at 60 mps)

T = 85 seconds( 1 min and 25 sec)

0 = 60+ 85a

Answer:

b- how much work can be done in a given time

how much work can be done in a given time

Answer:

60 kg

80 kg

Explanation:

Work is equal to the change in energy.

W = ΔE = E − E₀

Let's start with block B.  The work done by the tension force is equal to the change in energy.  Initially, the block has potential energy.  Finally, the block has kinetic energy.

W = ΔE

FΔy = ½ mv² − mgh

T (-2.0 m) = ½ m (6.00 m/s)² − m (10 m/s²) (2.0 m)

T (-2.0 m) = m (-2 m²/s²)

T = m (1 m/s²)

Now let's look at block A.  The work done by tension and against friction is equal to the change in energy.  Initially, the block has no energy.  Finally, it has both kinetic and potential energy.

W = ΔE

Fd = ½ mv² + mgh − 0

(T − Nμ) (2.0 m) = ½ (4.00 kg) (6.00 m/s)² + (4.00 kg) (10 m/s²) (⅗ × 2.0 m)

(T − Nμ) (2.0 m) = 120 J

T − Nμ = 60 N

Draw a free body diagram of block A and sum the forces in the perpendicular direction to find the normal force N.

N = mg cos θ

N = (4.00 kg) (10 m/s²) (⅘)

N = 32 N

Substitute:

T − 32μ = 60 N

If μ = 0, then T = 60 N and m = 60 kg.

If μ = ⅝, then T = 80 N and m = 80 kg.

Answer:did you find you answer also i answered so you can give him brainiest

Explanation:

Answer:

Impact Craters.

Explanation:

An impact crater can be defined as a circular depression that is caused by impact on any planet or asteroids or any other celestial body's surface. When smaller body in galaxy impacts these larger bodies, they form a circular depression on it's surface.

This is a major feature found in solid object such as the Moon, Mercury, etc.

Therefore, the feature that a planet does not require is an impact crater. As other features are important to define a planet. Thus correct option is C.

Answer:

103.52 km/h

Explanation:

We are given;

Distance between two cities; d = 4300 km

airspeed; v_as = 930 km/h

Difference in flight time; Δt = 61 min = 1.0167 h

Now, the equation of motion to find the distance is given as;

d = vt

Where v = v_as + v_js

v_as is the airspeed

v_js is the jet speed

Thus;

d = (v_as + v_js)t

Thus, time(t1) for outgoing flight is;

t1 = d/(v_as + v_js)

Meanwhile, time(t2) for the return flight, the jet stream velocity will be negative and time is;

t2 = d/(v_as - v_js)

Recall that Difference in flight time; Δt = 61 min.

Thus;

Δt = t2 - T1 = [d/(v_as - v_js)] - [d/(v_as + v_js)]

Factorizing out, we have;

Δt = d[1/(v_as - v_js)] - [1/(v_as + v_js)]

Furthermore, it gives;

Δt = d[(v_as + v_js - v_as + v_js)]/((v_as - v_js) × (v_as + v_js))

Δt = d(2v_js)/((v_as)² - (v_js)²)

Cross multiply to get;

(2dv_js)/Δt = ((v_as)² - (v_js)²)

(v_js)² + ((2dv_js)/Δt) - (v_as)² = 0

Plugging in values for d,v_as and Δt gives;

(v_js)² - ((2 × 4300 × v_js)/1.0167) - (930)² = 0

(v_js)² - (8458.7329v_js) - 864900 = 0

Using quadratic formula, we have;

v_js = 103.52 km/h

The airline computer advises an airspeed of 930 km/h, for which the difference in flight times for the outgoing and return flights is 61 min.  The jet stream speed obtained using the quadratic formula is  103.52 km/h.

Given:

Distance between two cities; d = 4300 km

airspeed; v(as) = 930 km/h

Difference in flight time; Δt = 61 min = 1.0167 h

Now, the equation of motion to find the distance is given as;

d = vt

Where v = v(as) + v(js)

d = v(as) + v(js)t

At time t₁, t₁ = d/v(as) + v(js)

At time t₂, t₂ = d/v(as) + v(js)

Recall that Difference in flight time; Δt = 61 min.

Δt =  d/v(as) + v(js) - d/v(as) + v(js)

Δt = d[(v(as) + v(js) - v(as) + v(js)]/((v(as) - v(js)) × (v(as) + v(js)))

Δt = d(2v(js))/((v(as))² - (v(js))²)

(2dv(js))/Δt = ((v(as))² - (v(js))²)

(v(js))² + ((2dv(js))/Δt) - (v(as))² = 0

(v(js))² - ((2 × 4300 × v(js))/1.0167) - (930)² = 0

(v(js))² - (8458.7329v(js)) - 864900 = 0

Using the quadratic formula, we have;

v(js) = 103.52 km/h

Therefore, The jet stream speed obtained using the quadratic formula is  103.52 km/h.

To know more about the quadratic formula:

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Explanation:

When the skater is dropped onto the ramp from above, the potential energy decreases and the kinetic energy increases.

Every time the skater bounces from the impact, thermal energy is gained, and both potential and kinetic energy are lost.

Answer:

whats the question?

Explanation:

Answer:

2 Key Concepts Heating and cooling can cause materials to change characteristics, such as state, color, and texture. Heating causes ice to become liquid water and cooling causes condensation to form on a window, mirror, or on the outside of a glass of water.

Explanation:

Answer:

a

   12

b

41667

Explanation:

From the question we are told that

Generally 1 dozen of egg is equivalent to 12 pieces of eye

So given that a gross of item is equivalent to 144 items

x dozens of eggs is equivalent to a gross of item

So

[tex]x = \frac{144 * 1}{12}[/tex]

=>     [tex]x  =  12 \ dozens \ of  \ eggs[/tex]

Generally the sum of  9260 and 32407 is  

               9 2 6 0

        +  3 2 4 0 7

           4  1  6 6 7

Answer:

Answer:

The object's mass.

Explanation:

The formula d=m/v.

d --> density

m--> mass

v --> volume

With density and volume given, you can calculate the mass of the object.

Explanation:

Answer:

Broadcasting is the method, not sure about the stage it is done in

Explanation:

Answer:

Broadcasting is the first (blank) second (blank) is Cultivation.

Explanation:

I took the test & got this answer correct.

Answer:

Average speed equals distance / time

380 / 40 = 9.5 m/s.

Explanation:

Answer:

A) v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]

B)√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]

C) µ_s = tan θ

D) µ_s = 0.4663

Explanation:

A) The forces acting on the car will be;

Force due to friction; F_f

Force due to Gravity; F_g

Normal Force; F_n

Now, let us take the vertical direction to be j^ and the direction approaching the centre to be i^ downwards and parallel to the road surface by k^.

Also, we will assume that initially, F_n is in the negative k^ direction and that it will have a maximum possible value of; F_f = µ_s × F_n

Thus, sum of forces about the vertical j^ direction gives;

ΣF_j^ = F_n•cos θ − mg + F_f•sin θ = 0

Since F_f = µ_s × F_n ;

F_n•cos θ − mg + (µ_s × F_n × sin θ) =0

F_n = mg/[cos θ + (µ_s•sin θ)]

Also, sum of forces about the centre i^ direction gives;

ΣF_i^ = F_n(sin θ + (µ_s•cos θ)) = mv²/r

Plugging in formula for F_n gives;

ΣF_i^ = [mg/[cos θ + (µ_s•sin θ)]] × (sin θ + (µ_s•cos θ)) = mv²/r

Making v the subject gives;

v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]

B) What we got in a above is the minimum speed the car can have while going round the turn.

The maximum speed will be gotten by making the frictional force(F_f) to point in the positive k^ direction. This means that F_f will be negative.

Now, if we change the sign in front of F_f in the equation in part a that led to the minimum velocity, we will have the maximum as;

v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]

Thus the range is;

√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]

C) For the minimum speed to be 0, it implies that F_f will be in the negative k^ direction. Thus, Sum of the forces in the k^ direction gives;

ΣF_k^ = mg(sin θ - µ_s•cos θ) = 0

Thus;

mg(sin θ - µ_s•cos θ) = 0

Making µ_s the subject gives;

µ_s = sin θ/cos θ

µ_s = tan θ

D) If θ = 25.0°;

Thus;

µ_s = tan 25

µ_s = 0.4663

Answer:

b) 0 = y₀ + 0 - ½ g t² ,  v₀ₓ = x / t

c)  v₀ₓ = 30.31 m / s

Explanation:

This is a projectile launching exercise, in your statement you have units of several systems, we are going to reduce everything to the SI system

   v = 35 mph (0.447 m / s / 1 mph) = 15.645 m / s

   y₀ = 75 m

A) In the attachment we can see a diagram of the movement of the vehicle in its fall

B) let's find the time for the fall to the base of the cliff (y = 0)

        y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

when the vehicle leaves the cliff it goes horizontally, so its vertical speed is zero (v_{oy} = 0)

         0 = y₀ + 0 - ½ g t²

         t = √2y₀ / g

with this time we can use the equation of motion on the x axis

         x = v₀ₓ t

         v₀ₓ = x / t

C) we perform the calculations

            t = √ (2 30 / 9.8)

            t = 2.474 s

           v₀ₓ = 75 / 2.474

           v₀ₓ = 30.31 m / s

D) as we can see, the vehicle speed when leaving the cliff is almost twice the allowed speed (15.6 m / s)

therefore the city is not responsible for the accident

The weld bead in SMAW is formed by the electrode wire core mixing with the molten base metal, Therefore the correct option is D.

It is a type of manual arc welding process which utilizes a consumable electrode covered with flux to protect the weld bead.

For SMAW either direct current or alternating current can be used to form the electric arc. The electrode melts and forms a molten weld pool with the base metal by using the thermal energy produced from the electric arc.

The weld bead in Shielded Metal Arc Welding is formed by the electrode wire core mixing with the molten base metal.

learn more about Shielded Metal Arc Welding (SMAW)

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Answer:

The quartering act of 1765 required the colonies to house British soldiers in barracks provided by the colonies.

Answer:

hope it helps...

Explanation:

The Principle of Moments states that when a body is balanced, the total clockwise moment about a point equals the total anticlockwise moment about the same point.

Answer:

The Principle of Moments states that when a body is balanced, the total clockwise moment about a point equals the total anticlockwise moment about the same point.

Answer:

Reduction division: The first cell division in meiosis, the process by which germ cells are formed. In reduction division, the chromosome number is reduced from diploid (46 chromosomes) to haploid (23 chromosomes). Also known as first meiotic division and first meiosis.

Explanation:

Please mark as brainliest

Answer:

Negative

Explanation:

duh

Answer:

:)

Explanation:

negative.

go add the snap carmel.bratz

Answer:

v = - 1,715 m / s ,   x = 0.0156 m

Explanation:

This is an oscillatory movement exercise, which is described by the expression

          x = A cos (wt + Ф)

we can assume that the block is released from its maximum elongation, so the phase constant (Ф) is zero

As we are told that the stone does not affect the movement of the spring mass system, the amplitude and angular velocity do not change, in the upward movement the stone is attached to the mass, but in the downward movement the mass has an acceleration greater than g leave the stone behind, let's look for time, for this we use the definition of speed and acceleration

         v = dx / dt

         v = - A w sin wt

          a = - Aw² cos wt

         a = -g

         -g = - Aw² cos wt

          wt = cos⁻¹ (g / Aw²)

          t = 1 / w cos⁻¹ (g / Aw²)

angular velocity and frequency are related

        w = 2π f

        w = 2π 4

         w = 8π   rad / s

remember that the angles are in radians

          t = 1 / 8π cos⁻¹ (9.8 / (0.07 64π²))

          t = 0.039789 1.3473

          t = 0.0536 s

let's find the speed for this time

          v = - A w sin wt

          v = - 0.07 8π sin (8π 0.0536)

          v = - 1,715 m / s

the distance is

           x = A cos wt

           x = 0.07 cos (8π 0.0536)

           x = 0.0156 m

Answer:

-time it takes for the sled to come to a stop after launch of rocket = 7.244 s

-distance sled has travelled from its starting point by the time it finally comes to rest is = 234.8655 m

Explanation:

From the question, looking at the motion while accelerating, we have;

Initial velocity; u = 0 m/s

Acceleration; a = 13.5 m/s²

Time; t = 3.3 s

Let's use first equation of motion to find final velocity (v).

v = u + at

v = 0 + (13.5 × 3.3)

v = 44.55 m/s

In this forward direction, let's calculate the displacement(d1) using newton's 3rd equation of motion.

d1 = ut + ½at²

d1 = 0(3.3) + ½(13.5 × 3.3²)

d1 = 73.5075 m

Now, let's consider the motion while slowing down and our final velocity will be 0 m/s while initial velocity will now be 44.55 m/s while acceleration is 6.15 m/s².

Thus, from v = u + at, we can find the time it take for the sled to come to a stop.

Now, since it's coming to rest acceleration will be negative. Thus;

0 = 44.55 + (-6.15t)

0 = 44.55 - 6.15t

t = 44.55/6.15

t = 7.244 s

Now we want to find out how far the sled has travelled from its starting point by the time it finally comes to rest.

Thus, we'll use the equation;

v² = u² + 2as

Where s will be the second displacement which we will call d2.

Thus;

0² = 44.55² + (-2 × 6.15 × s)

0 = 1984.7025 - 12.3s

12.3s = 1984.7025

s = 1984.7025/12.3

s = 161.358

Thus, d2 = s = 161.358 m

Thus, distance sled has travelled from its starting point by the time it finally comes to rest is ;

= d1 + d2 = 73.5075 + 161.358 = 234.8655 m