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3 years ago

A car is stopped at a red light. When the light turns green, it accelerates up

Physics

1 answer:

Marat540 [252]3 years ago

6 0

Answer:

a = 2.5 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f} =v_{o} +a*t$

where:

Vf = final velocity = 25 [m/s]

Vo = initial velocity = 0 (star from the rest)

a = acceleration [m/s²]

t = time = 10 [s]

25 = 0 + (a*10)

a = 25/10

a = 2.5 [m/s²]

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The mass of a sample of sodium bicarbonate is 2. 1 kilograms (kg). There are 1,000 grams (g) in 1 kg, and 1 Times. 109 nanograms

slega [8]

2.1 kg of sodium bicarbonate is equal to the 2.1 x 10¹² ng of sample. Option D is correct.

Mass is the quantity of the substance in the body or object. The SI unit of mass is Kilogram.

There are other units of measure,

Milligram: 1 g is equal to the $\bold {10^3 \ mg}$

Micro-gram: 1 g is equal to $\bold {10^{6} \ \mu g}$

Nano-gram: 1 g is is equal to $\bold {10^{9} \ ng}$

First convert kg to gram,

Since, 1 Kg = 1000 g

2.1 kg = grams of sample

So,

Do the cross multiplication,

$\rm mass\ of\ sample = \dfrac {2.1\ kg \times 1000\ g }{ 1 kg}\\\\\rm mass\ of\ sample =2100 g$

Now, convert 2100 g to nano-grams

Since, 1 g = 1 x 10⁹ ng

2100 g = ng of sample

So,

Do the cross multiplication,

$\rm mass\ of\ sample = \dfrac {2100 g \times 1 \times 10^9 ng }{1\ g}\\\\\rm mass\ of\ sample = 2.1 \times 10^1^2 ng$

Therefore, 2.1 kg of sodium bicarbonate is equal to the 2.1 x 10¹² ng of sample.

To know more about Mass units,

brainly.com/question/489186

3 0

3 years ago

a girl rides a sled down a frozen hill. she starts from rest at the top of the hill and reaches a speed of 16 m/s at the bottom

Zolol [24]

13.06m

Explanation:

Given parameters:

initial velocity u = 0

final velocity v = 16m/s

Unknown;

height of the hill = ?

Solution:

We are going to apply one of the laws of motion to solve this problem;

V² = U² + 2gH

where V is the final velocity

U is the initial velocity

g is the acceleration due to gravity = 9.8m/s²

H is the height

since initial velocity is zero;

V² = 2gH

To find the height;

H = $\frac{V^{2g} }{y}$

H = $\frac{16^{2} }{2 x 9.8}$ = 13.06m

learn more:

Velocity brainly.com/question/2706228

#learnwithBrainly

5 0

3 years ago

When a surface is illuminated with electromagnetic radiation of wavelength 480 nm, the maximum kinetic energy of the emitted ele

algol [13]

Answer:

Max kinetic energy for 340 nm wavelength will be $2.238\times 10^{-19}j$

Explanation:

In first case wavelength of electromagnetic radiation $\lambda =480nm=480\times 10^{-9}m$

Plank's constant $h=6.6\times 10^{-34}J-s$

Maximum kinetic energy = 0.54 eV

Energy is given by $E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{480\times 10^{-9}}=4.125\times 10^{-19}J$

We know that energy is given

$E=K_{MAX}+\Phi$ , here $\Phi$ is work function

So $4.125\times 10^{-19}=0.54\times 10^{-19}+\Phi$

$\Phi =3.585\times 10^{-19}J$

Now wavelength of second radiation = 340 nm

So energy $E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{340\times 10^{-9}}=5.823\times 10^{-19}J$

So $K_{MAX}=5.823\times 10^{-19}-3.585\times 10^{-19}=2.238\times 10^{-19}j$

6 0

4 years ago

Rigid Body Statics in 3 Dimensions

slamgirl [31]

Explanation:

Draw a free body diagram of the bar.

There are 3 reaction forces at O in the x, y, and z direction (Ox, Oy, and Oz).

There is a tension force Tac at A in the direction of the rope. There are also tension forces Tbd and Tbe at B in the direction of the ropes.

Finally, there is a weight force mg pulling down halfway between A and B, where m = 400 kg.

There are 6 unknown variables, so we'll need 6 equations to solve. Summing the forces in the x, y, and z direction will give us 3 equations. Summing the torques about the x, y, and z axes will give us 3 more equations.

First, let's find the components of the tension forces.

Tbe is purely in the z direction.

Tbd has components in the y and z directions. The length of Tbd is √8.

(Tbd)y = 2/√8 Tbd

(Tbd)z = 2/√8 Tbd

Tac has components in the x, y, and z directions. The length of Tac is √6.

(Tac)x = 1/√6 Tac

(Tac)y = 1/√6 Tac

(Tac)z = 2/√6 Tac

Sum of the forces in the +x direction:

∑F = ma

Ox − (Tac)x = 0

Ox − 1/√6 Tac = 0

Sum of the forces in the +y direction:

∑F = ma

Oy + (Tac)y + (Tbd)y − mg = 0

Oy + 1/√6 Tac + 2/√8 Tbd − mg = 0

Sum of the forces in the +z direction:

∑F = ma

Oz − (Tac)z − (Tbd)z − Tbe = 0

Oz − 2/√6 Tac − 2/√8 Tbd − Tbe = 0

Sum of the torques counterclockwise about the x-axis:

∑τ = Iα

mg (2 m) − (Tac)y (2 m) − (Tbd)y (2 m) = 0

mg − (Tac)y − (Tbd)y = 0

mg − 1/√6 Tac − 2/√8 Tbd = 0

Sum of the torques counterclockwise about the y-axis:

∑τ = Iα

-(Tac)x (2 m) + (Tbd)z (1.5 m) + Tbe (1.5 m) = 0

-4 (Tac)x + 3 (Tbd)z + 3 Tbe = 0

-4/√6 Tac + 6/√8 Tbd + 3 Tbe = 0

Sum of the torques counterclockwise about the z-axis:

∑τ = Iα

-mg (0.75 m) + (Tbd)y (1.5 m) = 0

-mg + 2 (Tbd)y = 0

-mg + 4/√8 Tbd = 0

As you can see, by summing the torques about axes passing through O, we were able to write 3 equations independent of those reaction forces. We can solve these equations for the tension forces, then go back and find the reaction forces.

-mg + 4/√8 Tbd = 0

4/√8 Tbd = mg

Tbd = √8 mg / 4

Tbd = √8 (400 kg) (9.8 m/s²) / 4

Tbd = 2772 N

mg − 1/√6 Tac − 2/√8 Tbd = 0

1/√6 Tac = mg − 2/√8 Tbd

Tac = √6 (mg − 2/√8 Tbd)

Tac = √6 ((400 kg) (9.8 m/s²) − 2/√8 (2772 N))

Tac = 4801 N

-4/√6 Tac + 6/√8 Tbd + 3 Tbe = 0

3 Tbe = 4/√6 Tac − 6/√8 Tbd

Tbe = (4/√6 Tac − 6/√8 Tbd) / 3

Tbe = (4/√6 (4801 N) − 6/√8 (2772 N)) / 3

Tbe = 653 N

Now, using our sum of forces equations to find the reactions:

Ox − 1/√6 Tac = 0

Ox = 1/√6 Tac

Ox = 1/√6 (4801 N)

Ox = 1960 N

Oy + 1/√6 Tac + 2/√8 Tbd − mg = 0

Oy = mg − 1/√6 Tac − 2/√8 Tbd

Oy = (400 kg) (9.8 m/s²) − 1/√6 (4801 N) − 2/√8 (2772 N)

Oy = 0 N

Oz − 2/√6 Tac − 2/√8 Tbd − Tbe = 0

Oz = 2/√6 Tac + 2/√8 Tbd + Tbe

Oz = 2/√6 (4801 N) + 2/√8 (2772 N) + 653 N

Oz = 6533 N

7 0

3 years ago

I need help. <br> How do I fill this out?

enyata [817]

Answer:

maybe try searching it up

6 0

3 years ago

Read 2 more answers