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3 years ago

How could the combustibility of a substance influence how the substances used

Physics

1 answer:

aliina [53]3 years ago

3 0

Answer:

Combustibility is a measure of how easily a substance bursts into flame, through fire or combustion. This is an important property to consider when a substance is used for construction or is being stored. It is also important in processes that produce combustible substances as a by-product.

Explanation:

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2.List two changes an unbalanced force can cause.

Evgen [1.6K]

Answer:

By applying an unbalanced force, you can change the motion of an object. Unbalanced forces can make an object at rest start moving, make a moving object stop, or change the direction and speed of the object.

Explanation:

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3 years ago

The speed of a boat undergoing constant acceleration increases from 5 meters per second to 15 meters per second over a period of

defon

Should be 250m. check with your teacher and let me know

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3 years ago

How do you find the total magnification of a microscope?

Alla [95]

In order to find total magnification of a microscope, you need to multiply the power of eyepiece and objective lens.

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4 years ago

Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t

Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

$v^2=v_0^2+2ax\\$

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

= 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

$a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}$

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

$F_f=ma=(1490kg)(5.55m/s^2)=8271.15N$

Furthermore, you use the relation between the friction force and the friction coefficient:

$F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56$

hence, the friction coefficient is 0.56

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3 years ago

According to Newton's Law of Cooling, if a body with temperature T 1 is placed in surroundings with temperature T 0, different f

Mila [183]

We can substitute the given values into the equation for T, given the surrounding temperature T0 = 0, initial temperature T1 = 140, constant k = -0.0815, and time t = 15 minutes.

T = 0 + (140 - 0)e^(-0.0815*15) = 140e^(-1.2225) = 41.23°F

3 0

3 years ago

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