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4 years ago

7

Select the correct answer.

2 answers:

Marrrta [24]4 years ago

5 0

Answer:

Matthew should use the safety glasses as his PPE.

Explanation:

nekit [7.7K]4 years ago

5 0

The answer is E. safety glasses

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the process of making alloys involves _____ pure metals to remove impurities. Then the pure metals are ____ with other component

Law Incorporation [45]

I think this is the answer for the first line(Cooling ,Heating or mixing ) and for the second line is(broken down,cooled,mixed)

5 0

3 years ago

Read 2 more answers

Which factors affect the strength of the electric force between two objects

Margaret [11]

<span>-- the product of the net charges on the objects;. -- the distance between the centers of their net charges. (Pretty much identical to the formula for gravitational force)</span>

8 0

3 years ago

Read 2 more answers

A 1,000 kg car is driving on a 15 m high bridge at 5 m/s. What is the kinetic energy of the car?

yanalaym [24]

Answer:

$KE=12,500J$

Explanation:

The formula for kinetic energy is:

$KE = \frac{1}{2}mv^2$

We can plug in the given values into the equation:

$KE = \frac{1}{2}*1000kg*(5m/s)^2$

$KE = 500kg*25m^2/s^2$

$KE=12,500J$

7 0

3 years ago

A block of mass 57.1 kg rests on a slope having an angle of elevation of 28.3°. If pushing downhill on the block with a force ju

stira [4]

Answer:

The coefficient is 0.90

Explanation:

Drawing a diagram makes thing easier, we will assume that the acceleration tends to zero because it start barely moving.

$-F_s+mg*sin(\theta)+F=0\\F_s=57.1kg*9.8m/s^2*sin(28.3)+177N\\F_s=442N\\F_s=\µ*N\\N=m*g*cos(\theta)\\N=57.1*9.8*cos(28.3)=493N\\\\\µ=\frac{442N}{493N}=0.90$

3 0

3 years ago

A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axles. A light cord wrapped around

Alex_Xolod [135]

Answer:

$\alpha =\frac{m*g*R}{I-m*R^2}$

$a = \frac{m*g*R^2}{I-m*R^2}$

$T=\frac{I*m*g}{I-m*R^2}$

Explanation:

By analyzing the torque on the wheel we get:

$T*R=I*\alpha$ Solving for T: $T=I/R*\alpha$

On the object:

$T-m*g = -m*a$ Replacing our previous value for T:

$I/R*\alpha-m*g = -m*a$

The relation between angular and linear acceleration is:

$a=\alpha*R$

So,

$I/R*\alpha-m*g = -m*\alpha*R$

Solving for α:

$\alpha =\frac{R*m*g}{I+m*R^2}$

The linear acceleration will be:

$a =\frac{R^2*m*g}{I+m*R^2}$

And finally, the tension will be:

$T =\frac{I*m*g}{I+m*R^2}$

These are the values of all the variables: α, a, T

8 0

3 years ago