How much distance did this object travel in meters between 0 and 10 seconds

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:

Explanation:

The density of a substance can be found by using the formula

[tex]density = \frac{mass}{volume} \\[/tex]

From the question

mass = 100 g

volume = 50 mL

We have

[tex] density = \frac{100}{50} = \frac{10}{5} \\ [/tex]

We have the final answer as

Hope this helps you

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:

1- the rate of flow of charge through a conductor is called current .whereas ,current density is the current per unit area of conductor

Answer:

D. The liquid is a mixture because it has a gas dissolved in a liquid.

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:

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:

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:

  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:

478.75 J

Explanation:

W=force* displacement

constant speed= (a=0) net F=0

Horizontal component of tension

Tcosx

125Ncos40= 95.76 N

W= (95.76 N)(5 m)

=478.75 J

The work done in moving the crate across the given distance is 478.75 J.

The given parameters;

The work done in moving the crate is the product of the horizontal component of the tension and the distance through which the crate is moved.

The work-done in moving the crate is calculated as;

W = Tcos(Ф) x d

W = 125cos(40) x 5

W = 478.75 J.

Thus, the work done in moving the crate across the given distance is 478.75 J.

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Answer:

It would take 5 seconds

Explanation:

I can't find the 'delta' sign nor the vector sign so just pretend that displacement, time and velocity has them.

V = d / t

34 = 170 / t

34 x t = 170

34t = 170

t = 170 / 34

t = 5

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

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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Answer:muscle strength

Explanation: bc well you would need it

Answer:

Negative

Explanation:

duh

Answer:

:)

Explanation:

negative.

go add the snap carmel.bratz

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:

A south pole attracts a north pole :)

Explanation:

Answer:

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

how much work can be done in a given time

Answer:

it gets colder

Explanation:

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

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:

When the mouse broke free from the hawk, its vertical velocity is zero since the mouse just fell from the grips of the hawk

We are given:

initial vertical velocity (v) = 0 m/s

initial horizontal velocity (u) = 10.1 m/s

height from the ground (h) = 15 m

acceleration due to gravity (a) = 10 m/s/s

final vertical velocity (vf) = v m/s

Solving for Final vertical velocity:

from the third equation of motion

(vf)² - (v)² = 2ah

replacing the variables

v² - 0² = 2(10)(15)

v² = 300

v = √300

v = 10√3     OR    17.3 m/s

Answer:

Explanation: someone help me please you want free points right

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.

The question is missing parts. The complete question is here.

The human circulatory system is closed, that is, the blood pumped out of the left ventricle of the heart into the arteries is constrained to a series of continuoous, branching vessels as it passes through the capillaries and then into the veins as it returns to the heart. The blood in each of the heart's four chambers comes briefly to rest before it is ejected by cotraction of the heart muscle. If the aorta (diameter [tex]d_{a}[/tex]) branches into two equal-sized arteries with a combined area equal to that of the aorta, what is the diameter of one of the branches?

(a) [tex]\sqrt{d_{a}}[/tex]

(b) [tex]\frac{d_{a}}{\sqrt{2} }[/tex]

(c) [tex]2d_{a}[/tex]

(d) [tex]\frac{d_{a}}{2}[/tex]

Answer: (b) [tex]\frac{d_{a}}{\sqrt{2} }[/tex]

Explanation: The cross-sectional area of a vessel is a circle. So area is:

[tex]A=\pi.r^{2}[/tex]

Radius is half of a diameter, i.e.:

[tex]r=\frac{d}{2}[/tex]

Suppose [tex]d_{b}[/tex] is diameter of the branches, radius of one the branches is:

[tex]r_{b}=\frac{d_{b}}{2}[/tex]

Both branches are equal sized, which means, both have the same radius. So, combined area of both branches is:

[tex]A_{b}=\pi\frac{d_{b}^{2}}{4} +\pi\frac{d_{b}^{2}}{4}[/tex]

[tex]A_{b}=2\frac{d_{b}^{2}}{4}[/tex]

Area of aorta is

[tex]A_{a}=\pi.(\frac{d_{a}}{2})^{2}[/tex]

[tex]A_{a}=\pi.\frac{d_{a}^{2}}{4}[/tex]

Area of aorta is equal the combined area of the branches, then:

[tex]2\pi\frac{d_{b}^{2}}{4}= \pi\frac{d_{a}^{2}}{4}[/tex]

Rearraging:

[tex]d_{b}^{2}=\frac{d_{a}^{2}}{2}[/tex]

[tex]d_{b}=\frac{d_{a}}{\sqrt{2} }[/tex]

The diameter of one of the branches is [tex]\frac{d_{a}}{\sqrt{2}}[/tex].

Answer:

The final speed of the second package is twice as much as the final speed of the first package.

Explanation:

Free Fall Motion

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

[tex]v=gt[/tex]

And the distance traveled downwards is:

[tex]\displaystyle y=\frac{gt^2}{2}[/tex]

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

[tex]\displaystyle t=\sqrt{\frac{2y}{g}}[/tex]

Replacing into the first equation:

[tex]\displaystyle v=g\sqrt{\frac{2y}{g}}[/tex]

Rationalizing:

[tex]\displaystyle v=\sqrt{2gy}[/tex]

Let's call v1 the final speed of the package dropped from a height H. Thus:

[tex]\displaystyle v_1=\sqrt{2gH}[/tex]

Let v2 be the final speed of the package dropped from a height 4H. Thus:

[tex]\displaystyle v_2=\sqrt{2g(4H)}[/tex]

Taking out the square root of 4:

[tex]\displaystyle v_2=2\sqrt{2gH}[/tex]

Dividing v2/v1 we can compare the final speeds:

[tex]\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}[/tex]

Simplifying:

[tex]\displaystyle v_2/v_1=2[/tex]

The final speed of the second package is twice as much as the final speed of the first package.