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KIM [24]
2 years ago
11

I need heelp can anyone heelp me plz

Physics
1 answer:
GuDViN [60]2 years ago
7 0

Answer:

e) 1.04

Explanation:

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What are the similarities between a car at rest and a car at a constant velocity
faltersainse [42]
They are both in motion because an object is not at rest, but moving so slow it could be at rest. A car going at the same constant velocity is neither speeding up or slowing down, an object "at rest" is also moving at a constant rate, not speeding up or slowing done.
8 0
3 years ago
A phonograph record accelerates from rest to 28.0 rpm in 5.73 s.
Arturiano [62]

Answer:

a) \alpha=0.5117\ rad.s^{-2}

b) \theta=8.4\ rad

Explanation:

Given:

  • initial rotational speed of phonograph, \omega_i=0\ rad.s^{-1}
  • final rotational speed of phonograph, N_f=28\ rpm \Rightarrow \omega_f=2\pi\times\frac{28}{60} =2.932\ rad.s^{-1}
  • time taken for the acceleration, t=5.73\ s

a)

Now angular acceleration:

\alpha=\frac{\omega_f-\omega_i}{t}

\alpha=\frac{2.932-0}{5.73}

\alpha=0.5117\ rad.s^{-2}

b)

Using eq. of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=0+\frac{1}{2}\times 0.5117\times 5.73^2

\theta=8.4\ rad

5 0
3 years ago
You spill a bucket of soapy water on a marble floor accidentally.Would it make it easier or difficult for you to walk on the flo
stiks02 [169]

Answer:

Of course harder

Explanation:

Just imagone the floor is wet and you walk on it, do u feel it hard or easy to walk? :D

7 0
2 years ago
Read 2 more answers
What is the freezing point of an aqueous solution that boils at 101 degrees C? Kfp = 1.86 K/m and Kbp = 0.512 K/m. Enter your an
astra-53 [7]

Answer:

-3.63 degree Celsius

Explanation:

We are given that

Boiling point of solution=T_b=101^{\circ} C

Boiling point water=100 degree Celsius

K_f=1.86K/m

K_b=0.512 K/m

\Delta T_b=T-T_0

Where T=Boiling point of solution

T_0=Boiling point of pure solvent

\Delta T_b=101-100=1^{\circ}C

\Delta T_b=k_bm

Using the formula

1=0.512\times m

Molality,m=\frac{1}{0.512} m

\Delta T_f=k_fm

Using the formula

\Delta T_f=\frac{1}{0.512}\times 1.86

\Delta T_f=3.63 C

We know that

\Delta T_f=T_0-T_1

Where T_0 =Freezing point of solvent

T_1= Freezing point of solution

Using the formula

3.63=0-T_1

Freezing point of water=0 degree Celsius

T_1=0-3.63=-3.63 C

Hence, the freezing point of solution=-3.63 degree Celsius

7 0
2 years ago
A block of lead with dimensions 2.0 dm x 8.0 cm x 35 mm, has a mass of 6.356 kg. calculate the density of lead in g/cm
insens350 [35]

First, let us calculate for the volume of the block of lead using the formula:

V = l * w * h

But we have to convert all units in terms of cm:

l = 2.0 dm = 20 cm

w = 8 cm

h = 3.5 cm

 

Therefore the volume is:

V = (20 cm) * (8 cm) * (3.5 cm)
V = 560 cm^3

 

Next we convert the mass in terms of g:

m = 6.356 kg = 6356 g

 

Density is mass over volume, so:

density = 6356 g / 560 cm^3

density = 11.35 g / cm^3     (ANSWER)

8 0
3 years ago
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