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3 years ago

"the time range specified for a historical search defines the" _______________ .

1 answer:

5 0

The answer is period. This is the date range for which you ask the system to return the events depending on the search keys, and keywords you have entered.

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Two forces act on a 6.00-kg object. One of the forces is 10.0 N. If the object accelerates at 2.00 m/s2

liubo4ka [24]

Given :

Two forces act on a 6.00-kg object. One of the forces is 10.0 N.

Acceleration of object 2 m/s².

To Find :

The greatest possible magnitude of the other force.\

Solution :

Let, other force is f.

So, net force, F = 10 + f.

Now, acceleration is given by :

$a=\dfrac{F}{mass}\\\\a= \dfrac{10+f}{6}\\\\\dfrac{10+f}{6}=2\\\\f = 12 - 10\\\\f = 2 \ N$

Therefore, the greatest possible magnitude of the other force is 2 N.

Hence, this is the required solution.

7 0

2 years ago

In a low-density cloud, the internal pressure is much too low to support the weight, so the cloud must __________, eventually pr

andrey2020 [161]

The answer to this question is flatten

so the cloud must <u>flatten</u> eventually producing the galaxy s disk.

3 0

3 years ago

EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE AHHHHHHHHHHHHHHHHHHHHHHHHHHH COCOA PEBBLES RULE HEHEHEHEHEHEHE

Andrew [12]

Answer:

AAAAAAAAHHHDHSFJSBDKFANEDASNXDEHAHDBWEEEEHJDBHBAH

AAHHH YES THEY DOO AAAHHH EEHSFJWEHASAAAAHJEHER

7 0

3 years ago

Read 2 more answers

) In the figure below, three charges K, L and M are placed at the corners of an equilateral triangle. If K and L are fixed, in w

abruzzese [7]

Answer:

Explanation:

Negative (-) charge M will not move towards negative (-) charge K because, same charges will not attract each other in the given case

Negative (-) charge at the M tends to move towards positive (+) charge L in the direction of B) because opposite charges attract each other.

4 0

3 years ago

A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim

Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

$F_D = -\frac{1}{2}\rho V^2 C_d A$

Where

$\rho$ is the density of the flow

V = Velocity

$C_d$ = Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

$F=ma=m\frac{dV}{dt}$

Equating both equations we have:

$m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A$

$m(dV)=-\frac{1}{2}\rho C_d A (dt)$

$\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)$

Integrating

$\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)$

$-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t$

Here,

$V_f = 60mph = 26.82m/s$

$V_i = 120.7m/s$

$m= 1600lbf = 725.747Kg$

$\rho = 1.21 kg/m^3$

$C_d = 0.3$

$\Delta t=7s$

Replacing:

$\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)$

$-0.029 = -5.4997r^2$

$r = 2.2963m$

$d= r*2 = 4.592m \approx 15.065ft$

4 0

3 years ago