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Maksim231197 [3]

3 years ago

7

John doe gets on the highway in his 1967 Shelby 427 Cobra starting from the dead stop at the bottom of the on ramp of it can be

assumed that John accelerates at a rate of 6.7 m s2 how long does it take for John to reach the speed of 30 m s?

1 answer:

masha68 [24]3 years ago

7 0

Answer:

It will take john 4.477 seconds

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What happens to beaches over time?

Rama09 [41]

the answer is d i took the test be cause i go to a k12 onlie school

7 0

3 years ago

Read 2 more answers

Holding onto a tow rope moving parallel to a frictionless ski slope, a 70.1 kg skier is pulled up the slope, which is at an angl

lara [203]

Answer:

given,

mass of the skier = 70.1 Kg

angle with horizontal, θ = 8.6°

magnitude of the force,F = ?

a) Applying newton's second law

$m g sin\theta - F_{rope} = ma$

velocity is constant, a = 0

$m g sin\theta - F_{rope} =0$

$F_{rope} = m g sin\theta$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0$

$F_{rope}= 102.73\ N$

b) now, when acceleration, a = 0.135 m/s²

$F_{rope}-m g sin\theta = ma$

velocity is constant, a = 0.135 m/s₂

$F_{rope} = m g sin\theta+ma$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0+70.1\times 0.135$

$F_{rope}= 112.19\ N$

7 0

3 years ago

Isaac Newton's first law of motion states?

lesya [120]

Answer:

His first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. In other words its inertia.

Explanation:

8 0

3 years ago

The temperature and time t given in hours from 0 to 24 after midnight in downtown mathville is given by t=10-5 sin(pi/12 t) degr

Dvinal [7]

Answer:

$T_A_v_g=9.918192559$^{\circ}C$

Explanation:

The problem tell us that the temperature as function of time in downtown mathville is given by:

$T(t)=10-5*sin(\frac{\pi}{12t})$

The average temperature over a given interval can be calculated as:

$T_a_v_g=\frac{T_o+T_f}{2}$

Where:

$T_o=Initial\hspace{3}temperature\\T_f=Final\hspace{3}temperature$

So, the initial temperature in this case, would be the temperature at noon, and the final temperature would be the temperature at midnight:

Therefore:

$T_o=T(12)=10-5*sin(\frac{\pi}{12*12}) =9.890925575$^{\circ}C$

$T_f=T(24)=10-5*sin(\frac{\pi}{12*24}) =9.945459543$^{\circ}C$

Hence, the average temperature between noon and midnight is:

$T_A_v_g=\frac{9.890925575+9.945459543}{2}=9.918192559$^{\circ}C$

3 0

3 years ago

The rate (in cubic feet per hour) that a spherical snowball melts is proportional to the snowball's volume raised to the 2/3 pow

Darina [25.2K]

Answer:

A 3 feet radius snowball will melt in 54 hours.

Explanation:

As we can assume that the rate of snowball takes to melt is proportional to the surface area, then the rate for a 3 feet radius will be:

T= A(3 ft)/A(1 ft) * 6 hr

A is the area of the snowballs. For a spherical geometry is computing as:

A=4.pi.R^2

Then dividing the areas:

A(3 feet)/A(1 foot) = (4 pi (3 ft)^2)/(4 pi (1 ft)^2) = (36pi ft^2)/(4pi ft^2)= 9

Finally, the rate for the 3 feet radius snowball is:

T= 9 * 6 hr = 54 hr

6 0

3 years ago