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3 years ago

If I start driving 80 miles per hour down the interstate how far will I go if I Drive for 4.5 hours

1 answer:

4 0

If you are driving 80 miles per hour, you are driving 80 miles for every hour. So if you drive for 4.5 hours, multiply. 80 x 4.5= 360miles

helpful formula:S = V x T

Speed= Velocity(80mph) x Time(4.5hrs)

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What are used for manufacturing paper along with chemical

Airida [17]

Answer:

Dry-strength additives, or dry-strengthening agents, are chemicals that improve paper strength normal conditions. These improve the paper's compression strength, bursting strength, tensile breaking strength, and delamination resistance. Typical chemicals used include cationic starch and polyacrylamide derivatives.

8 0

2 years ago

A small object with a 5.0-mC charge is accelerating horizontally on a friction-free surface at 0.0050 m/s2 due only to an electr

kolbaska11 [484]

Answer:

$0.002 N/C$

Explanation:

Parameters given:

Charge of object, q = 5 mC = $5 * 10^{-3} C$

Acceleration of object, a = $0.005 m/s^2$

Mass of object, m = 2.0 g

The Electric field exerts a particular force on the object, causing it to accelerate (Electrostatic force).

We know that Electrostatic force, F, is given in terms of Electric field, E, as:

F = qE

This means that the object exerts a force of -qE on the Electric force (Action with equal and opposite reaction).

The object also has a force, F, due to its acceleration a. This force is the product of its mass and acceleration. Mathematically:

F = ma

Equating the two forces of the object, we get:

-qE = ma

=> $E = \frac{-ma}{q}$

Solving for E, we have:

$E = \frac{-2 * 10^{-3} * 0.005}{5 * 10^{-3}} \\\\\\E = -0.002 N/C$

The magnitude will be:

$|E| = |-0.002| N/C = 0.002 N/C$

The electric field has a magnitude of 0.002 N/C.

4 0

3 years ago

A 25.0-g sample of copper at 363 K is placed in I 00.0 g of water at 293 K. The copper and water quickly come to the sa me tempe

Simora [160]

Answer:

Final temperature is 295K

Explanation:

Where the sample of copper is placed in the water, the heat transferred from the copper is equal that the heat absorbed by the water.

The heat transferred from the copper is:

C× $\frac{1mol}{63,546g}$ ×mass×ΔT

Where C is molar heat capacity of copper (24,5J/molK)

Mass is 25,0g

And ΔT is final temperature - initial temperature (X-363K)

Also, the heat absorbed by the water is:

-C× $\frac{1mol}{18,02g}$ ×mass×ΔT

Where C is molar heat capacity of water (75,2J/molK)

Mass is 100,0g

And ΔT is final temperature - initial temperature (X-293K)

As heat transferred is equal to heat absorbed:

24,5J/molK× $\frac{1mol}{63,546g}$ ×25,0g×(X-363K) = -75,2J/molK× $\frac{1mol}{18,02g}$ ×100,0g× (X-293K)

9,64X J/K - 3499J = - 417X J/K + 122273J

426,64X J/K = 125772 J

X = 295K

Final temperature is 295K

I hope it helps!

6 0

3 years ago

The moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is

sleet_krkn [62]

Answer:

I = I₀ + M(L/2)²

Explanation:

Given that the moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is I₀.

The parallel axis theorem for moment of inertia states that the moment of inertia of a body about an axis passing through the centre of mass is equal to the sum of the moment of inertia of the body about an axis passing through the centre of mass and the product of mass and the square of the distance between the two axes.

The moment of inertia of the body about an axis passing through the centre of mass is given to be I₀

The distance between the two axes is L/2 (total length of the rod divided by 2

From the parallel axis theorem we have

I = I₀ + M(L/2)²

5 0

3 years ago

Step 2: Apply NEwton's second law Apply ∑Fy = may , what should ay be equal to, since the block doesn't move in the y direction

andrey2020 [161]

Answer:

∑Fy = 0, because there is no movement, N = m*g*cos (omega)

Explanation:

We can solve this problem with the help of a free body diagram where we show the respective forces in each one of the axes, y & x. The free-body diagram and the equations are in the image attached.

If the product of mass by acceleration is zero, we must clear the normal force of the equation obtained. The acceleration is equal to zero because there is no movement on the Y-axis.

3 0

3 years ago