Physical properties of minerals graphic organizer

A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force

Tamiku [17]

Answer:

(a) 91 kg (2 s.f.) (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

$s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}$

Subsequently,

$F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})$

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

$v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}$

According to Newton's First Law which states objects will remain at rest

or in uniform motion (moving at constant velocity) unless acted upon by

an external force. Hence, the block of ice by the end of the first 5

seconds, experiences no acceleration (a = 0) but travels with a constant

velocity of 4.4 $m \ s^{-1}$ .

$s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}$

Therefore, the ice block traveled 22 m in the next 5 seconds after the

worker stops pushing it.

4 0

2 years ago

A flute player hears four beats per second when she compares her note to an 880 Hz tuning fork (note A). She can match the frequ

ludmilkaskok [199]

Answer:

884Hz

Explanation:

Beats is the absolute difference between two frequencies therefore

Beats = f1-f2

4=f1-880

F1=880+4

F1=884Hz

7 0

3 years ago

True or false? Magnetic reversals are recorded in the newly formed oceanic crust on BOTH sides of a mid-ocean ridge spreading ce

andrey2020 [161]

answer: false is my answer for thi question

4 0

3 years ago

Calculate the acceleration of a 1000 kg car if the motor provides a small thrust of 1000 N and the static and dynamic friction c

grin007 [14]

Explanation :

It is given that,

Mass of the car, m = 1000 kg

Force applied by the motor, $F_A=1000\ N$

The static and dynamic friction coefficient is, $\mu=0.5$

Let a is the acceleration of the car. Since, the car is in motion, the coefficient of sliding friction can be used. At equilibrium,

$F_A-\mu mg=ma$

$\dfrac{F_A-\mu mg}{m}=a$

$a=\dfrac{1000-0.5(1000)(9.81)}{1000}$

$a=-3.905\ m/s^2$

So, the acceleration of the car is $-3.905\ m/s^2$ . Hence, this is the required solution.

6 0

3 years ago

A negative charge of -2.0 C and a positive charge of 3.0 C are separated by 80 m. What is the electrostatic force between the tw

faltersainse [42]

Answer:

1. 8437500 N

2. The force between the two charges is attractive.

Explanation:

1. Determination of the force between the two charges.

Charge 1 (q₁) = –2.0 C

Charge 2 (q₂) = 3.0 C

Distance apart (r) = 80 m

Electrical constant (K) = 9×10⁹ Nm²/C²

Force (F) =?

F = Kq₁q₂ / r²

F = 9×10⁹ × 2 × 3 / 80²

F = 5.4×10¹⁰ / 6400

F = 8437500 N

Thus, the force of attraction between the two charges is 8437500 N

2. From the question given, the charges are:

Charge 1 (q₁) = –2.0 C

Charge 2 (q₂) = 3.0 C

We understood that like charges repels while unlike charges attract. Since the two charges (i.e –2 C and 3 C) has opposite signs, it means they will attract each other.

Thus the force between them is attractive.

6 0

3 years ago