A table is moved using 60 N of force.

Is a tape dispenser a pulley or a wheel and axis?

Sindrei [870]

It is a wheel and axis

4 0

3 years ago

Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg

KengaRu [80]

Answer

given,

speed of car = 60 km/hr

= 60 x 0.278 m/s

= 16.68 m/s

radius of the tire = 25 cm = 0.25 m

time taken to stop = 6 s

average acceleration = ?

$\omega = \dfrac{v}{R}$

$\omega = \dfrac{16.68}{0.25}$

$\omega =66.72 rad/s$

average acceleration= $\dfrac{\Delta \omega}{\Delta t}$

= $\dfrac{66.72}{6}$

= 11.12 rad/s²

correct answer option D

b) length of sea saw = 10 m

mass of jack = 100 Kg

mass of Jill = 60 Kg

Let y be the distance from the Jack. To balance the seesaw moment on both side should balance each other about fulcrum

now,

100 x y = 5 x 60

y = 3 m

the correct answer is option B

4 0

3 years ago

What are many scientific laws represented with?

salantis [7]

hypothesises i think

3 0

4 years ago

72) What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 (formula weight = 164 g/mol) in 115

Rina8888 [55]

<u>Answer:</u> The freezing point of solution is -3.34°C

<u>Explanation:</u>

Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of calcium nitrate follows:

$Ca(NO_3)_2(aq.)\rightarrow Ca^{2+}(aq.)+2NO_3^-(aq.)$

The total number of ions present in the solution are 3.

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(Ca(NO_3)_2)$ = 11.3 g

$M_{solute}$ = Molar mass of solute $(Ca(NO_3)_2)$ = 164 g/mol

$W_{solvent}$ = Mass of solvent (water) = 115 g

Putting values in above equation, we get:

$\text{Molality of }Ca(NO_3)_2=\frac{11.3\times 1000}{164\times 115}\\\\\text{Molality of }Ca(NO_3)_2=0.599m$

To calculate the depression in freezing point, we use the equation:

$\Delta T=iK_fm$

where,

i = Vant hoff factor = 3

$K_f$ = molal freezing point depression constant = 1.86°C/m.g

m = molality of solution = 0.599 m

Putting values in above equation, we get:

$\Delta T=3\times 1.86^oC/m.g\times 0.599m\\\\\Delta T=3.34^oC$

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T=\text{freezing point of water}-\text{freezing point of solution}$

$\Delta T$ = 3.34 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

$3.34^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-3.34^oC$

Hence, the freezing point of solution is -3.34°C

4 0

4 years ago

A crow is flying horizontally with a constant speed of 2.70m/s when it releases a claim from its beak. The clan lands on the roc

jenyasd209 [6]

Given:

Speed = 2.70 m

Time, t = 2.10 seconds

Let's solve for the following:

• (a) The horizontal component of the velocity.

To find the horizontal component, apply the formula:

$V_{ox}=V_o\cos \theta$

Where:

Vo is the initial speed = 2.70 m

θ = 0 degrees

Hence, we have:

$\begin{gathered} V_{ox}=2.70\cos 0 \\ \\ V_{ox}=2.7\text{ m/s} \end{gathered}$

The horizontal component of the velocity just before it lands is 2.70 m/s.

• (b) The vertical component of the velocity.

To find the vertical component, apply the formula:

$V_{oy}=V_{0y}-gt=\text{V}_{oy}\text{ sin}\Theta-gt$

Where:

g is the acceleration due to gravity = 9.8 m/s²

t is the time = 2.10 s

Hence, we have:

$\begin{gathered} V_{oy}=V_{oy}\sin \theta-gt \\ \\ V_{oy}=2.70\sin 0-9.8(2.10) \\ \\ V_{oy}=0-20.58 \\ \\ V_{oy}=-20.58\text{ m/s} \end{gathered}$

The vertical component of the velocity just before it lands is -20.58 m/s.

(c) Here, the initial speed is equal to the constant horizontal speed.

Therefore, in part (a) the horizontal component will increase in the x-direction if the speed of the crow is increased.

The initial vertical velocity is 0 m/s in both cases.

Therefore, in part (b) the vertical component will remain constant.

ANSWER:

(a) 2.70 m/s

(b) -20.58

(c) In part (a) the horizontal component will increase, while in part (b) the vertical component will remain constant.

6 0

1 year ago