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creativ13 [48]
3 years ago
11

You want to slide a 0.39 kg book across a table. If the coefficient of kinetic friction is .21, what force is required to move t

he book with an acceleration of 0.18 m/s^2?
Physics
1 answer:
uranmaximum [27]3 years ago
5 0
Find the force that would be required in the absence of friction first, then calculate the force of friction and add them together.  This is done because the friction force is going to have to be compensated for.  We will need that much more force than we otherwise would to achieve the desired acceleration:

F_{NoFric}=ma=0.39kg \times0.18 \frac{m}{s^2}  =0.0702N

The friction force will be given by the normal force times the coefficient of friction.  Here the normal force is just its weight, mg

F_{Fric}=0.39kg \times 9.8 \frac{m}{s^2} \times 0.21=0.803N

Now the total force required is:

0.0702N+0.803N=0.873N

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The SI system uses three base units. Is this true or false?
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The\text{ SI }system\text{ uses seven }base\text{ units, hence the statement is false}

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1 year ago
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The velocity of the transverse waves produced by an earthquake is 8.9 km/s, and that of the longitudinal waves is 5.1 km/s. A se
Brrunno [24]

Answer: The distance is 723.4km

Explanation:

The velocity of the transverse waves is 8.9km/s

The velocity of the longitudinal wave is 5.1 km/s

The transverse one reaches 68 seconds before the longitudinal.

if the distance is X, we know that:

X/(9.8km/s) = T1

X/(5.1km/s) = T2

T2 = T1 + 68s

Where T1 and T2 are the time that each wave needs to reach the sesmograph.

We replace the third equation into the second and get:

X/(9.8km/s) = T1

X/(5.1km/s) = T1 + 68s

Now, we can replace T1 from the first equation into the second one:

X/(5.1km/s) = X/(9.8km/s) + 68s

Now we can solve it for X and find the distance.

X/(5.1km/s) - X/(9.8km/s) = 68s

X(1/(5.1km/s) - 1/(9.8km/s)) = X*0.094s/km= 68s

X = 68s/0.094s/km = 723.4 km

6 0
3 years ago
Calculate the molecular weight of Aluminium hydroxide​
Vedmedyk [2.9K]

Al(OH)3 = 26.98 + [(16×3) + (1.01×3)] = 26.98 + 51.03 = 78.01 and the unit will be g/mol

<h3><em>Al(OH)3 = 78.01 g/mol</em></h3>
5 0
3 years ago
2. In the pendulum lab, the period (swing time) was O a. the independent variable O b.graphed on the vertical-axis O c.graphed o
Nadya [2.5K]
B is appropriate answer
7 0
3 years ago
To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0129-kg bullet i
ipn [44]

Answer:

t=0.42s

Explanation:

Here you have an inelastic collision. By the conservation of the momentum you have:

m_1v_1+m_2v_2=(m_1+m_2)v

m1: mass of the bullet

m2: wooden block mass

v1: velocity of the bullet

v2: velocity of the wooden block

v: velocity of bullet and wooden block after the collision.

By noticing that after the collision, both objects reach the same height from where the wooden block was dropped, you can assume that v is equal to the negative of v2. In other words:

m_1(-v_1)+m_2v_2=(m1+m2)(-v2)

Where you assumed that the negative direction is upward. By replacing and doing v2 the subject of the formula you get:

-(0.0129kg)(767m/s)+(1.17kg)v_2=(1.1829kg)(-v_2)\\\\v_2=4.20m/s

Now, with this information you can use the equation for the final speed of an accelerated motion and doing t the subject of the formula. IN other words:

v_2=v_o+gt\\\\t=\frac{v_2-v_o}{g}=\frac{4.2m/s-0m/s}{9.8m/s^2}=0.42s

hence, the time is t=0.42 s

4 0
3 years ago
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