2 years ago

Which of the following best represents a chemical reaction?

Physics

1 answer:

Sonja [21]2 years ago

7 0

Answer:

The answer to your question should be D.

Explanation:

reactants are on the laft side of arrow and products are on right side of arrow

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HELP I WILL GIVE BRAINLIEST!

Ann [662]

Answer:

Answer:B

Explanation:

Because it all stayed consistant

4 0

2 years ago

What is the current through a 11v bulb with a power of 99w

Murrr4er [49]

Answer:

9.01amp

Explanation:

Power = V^2/R

Given that v = 11volts, P = 99watts

99 = 11^2/R

11×11 = 99R

121= 99R

R = 121/99

R= 1.22ohms

From ohms Law; V = IR

11volts = I × 1.22ohms

I = 11/1.23

I = 9.01 amp

8 0

3 years ago

Which is the mode of the following data?<br>120, 125, 130, 125, 135

Aleks [24]

The mode in this case would be 125 because it occurs the most in the sequence of numbers.

8 0

3 years ago

Read 2 more answers

The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the t

Nina [5.8K]

Answer:

The speed of the boxes are 1 m/s.

Explanation:

Given that,

Mass of box = 1 kg

Mass of another box = 2 kg

Suppose 1 kg box moves with 3 m/s speed.

We need to calculate the speed of the boxes

Using formula of conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v$

Where, u = initial velocity

v = final velocity

Put the value into the formula

$1\times3+2\times0=(1+2)v$

$v=\dfrac{3}{3}$

$v=1\ m/s$

Hence, The speed of the boxes are 1 m/s.

4 0

3 years ago

Occasionally, people can survive falling large distances if the surface they land on is soft enough. During a traverse of Eiger'

uranmaximum [27]

Answer:

4611.58 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.174 ft/s²

Equation of motion

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 32.174\times 516+0^2}\\\Rightarrow v=182.218\ ft/s$

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-182.218^2}{2\times 3.6}\\\Rightarrow a=-4611.58\ ft/s^2$

Magnitude of acceleration while stopping is 4611.58 ft/s²

8 0

2 years ago