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3 years ago

HELP WITH SCIENCE STAAR QUESTIONS

Physics

1 answer:

nikdorinn [45]3 years ago

5 0

Answer:

EASY PEASY LEMON-SQUEEZY.

Explanation:

Mark Brainliest.

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Three forces act on an object. Two of the forces have the magnitudes 56 N and 27 N, and make angles 53° and 156°, respectively,

Ludmilka [50]

Answer:

Explanation:

Given

Two forces $F_1$ and $F_2$ at an angle of $\theta _1$ $\theta _2$

$F_1=56\ N$

$F_2=27\ N$

$\theta _1=53^{\circ}$

$\theta _2=156^{\circ}$

As resultant force is zero therefore horizontal component as well as vertical component of force is zero

$\sum F_x=F_1\cos \theta _1+F_2\cos \theta _2+F_3\cos \theta _3$

$\sum F_x=56\cos 53+27\cos 156+F_3\cos \theta$

$F_3\cos \theta_3=-9.035\ N----1$

$\sum F_y=F_1\sin \theta _1+F_2\sin \theta _2+F_3\sin \theta _3$

$F_3\sin \theta _3=55.704\ N----2$

squaring and adding 1 and 2

$F_3^2(\cos ^2\theta _3+\sin^2\theta _3)=86.583+3102.93$

$F_3=56.47\ N$

Divide 1 and 2 to get $\theta _3$

$\frac{\sin \theta_3 }{\cos \theta_3 }=\frac{-55.704}{9.305}$

$\tan \theta_3=-6.16$

$\theta _3=180-80.78$

$\theta _3=99.22^{\circ}$

6 0

3 years ago

When all individual forces acting upon an object are balanced, it is the natural tendency of the object to

sleet_krkn [62]

Answer:

- maintain its state of motion

- Keep its velocity constant (either at zero or non-zero value)

8 0

3 years ago

Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug

Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

$F_D=\dfrac{1}{2}\rho C_DA v^2$

$C_D$ =Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

Now by putting the values

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

$\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}$

$\dfrac{F_D_1}{F_D_2}=0.444$

Percentage Change in the drag force

$\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100$

$\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100$

$\Delta F=\dfrac{1-0.444}{0.444}\times 100$

ΔF=125.22 %

Therefore force will increase by 125.22 %.

3 0

3 years ago

What is the average speed of particles of atoms at room temperature?

Paha777 [63]

Answer:

300 meters per second. That's equal to about 670 miles per hour.

Explanation:

Not only are air particles incredibly small, they are always moving. And they move fast. At room temperature, they are going about 300 meters per second. That's equal to about 670 miles per hour.

8 0

4 years ago

A 50.0 g toy car is released from rest on a frictionless track with a vertical loop of radius R (loop-the-loop). The initial hei

Mariana [72]

Answer:

the speed of the car at the top of the vertical loop $v_{top} = 2.0 \sqrt{gR \ \ }$

the magnitude of the normal force acting on the car at the top of the vertical loop $F_{N} = 1.47 \ \ N$

Explanation:

Using the law of conservation of energy ;

$mgh = mg (2R) + \frac{1}{2}mv^2_{top}\\\\mg ( 4.00 \ R) = mg (2R) + \frac{1}{2}mv^2_{top}\\\\g(4.00 \ R) = g (2R) + \frac{1}{2}v^2 _{top}\\\\v_{top} = \sqrt{2g(4.00R - 2R)}\\\\v_{top} = \sqrt{2g(4.00-2)R$

$v_{top} = 2.0 \sqrt{gR \ \ }$

The magnitude of the normal force acting on the car at the top of the vertical loop can be calculated as:

$F_{N} = \frac{mv^2_{top}}{R} \ - mg\\\\F_{N} = \frac{m(2.0 \sqrt{gR})^2}{R} \ - mg\\\\F_{N} = [(2.0^2-1]mg\\\\F_{N} = [(2.0)^2 -1) (50*10^{-3} \ kg)(9.8 \ m/s^2]\\\\$

$F_{N} = 1.47 \ \ N$

4 0

4 years ago