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2 years ago

A student is given a sample of matter and asked to determine whether it is an element.

Physics

2 answers:

Pepsi [2]2 years ago

5 0

I think the correct answer is C

scoundrel [369]2 years ago

4 0

It's a mixture cause the 3 different pure substances can only be separated if its a mixture

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) A physics student wants to measure the stiffness of a spring (force required per cm stretched). He knows that according to Hoo

Maurinko [17]

Answer:

Explanation:

find the solution below

3 0

2 years ago

A 3 mm inside diameter tube is placed in a fluid with a surface tension of 600 mN/m and density of 3.7 g/cm3. The contact angle

Aleks04 [339]

Answer: The height of the fluid rise is 0.01m

Explanation:

Using the equation

h = (2TcosѲ )/rpg

h= height of the fluid rise

diameter of the tube =3mm

radius of the tube= 3/2 =1.5mm=0.0015

T= surface tension = 600mN/m=0.6N/m

Ѳ = contact angle = $60^o$ C

p= density =3.7g/cm3= 3700kg/m3

g= acceleration due to gravity =9.8m/s2

h = ( 2*0.6*0.5)/(0.0015*3700*9.8)

h = 0.6/54.39

h= 0.01m

Therefore,the height of the fluid rise is 0.01m

8 0

2 years ago

A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle

tatiyna

Answer:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

(a) find speed v

vertical displacement

$-h= vsinA\times t-gt^2/2$

putting values h=15 m, v=0.8

$-15 = 0.8vt - 4.9t^2$ ............. (1)

horizontal displacement d = vcosA×t = 0.6×v ×t

so v×t = d/0.6 = 40/0.6

plug it into (1) and get

$-15 = 0.8\times40/0.6 - 4.9t^2$

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) = 17.8 m/s

(b) If his speed was only half the value found in (a), where did he land?

v = 17.8/2 = 8.9 m/s

vertical displacement = $-H =v sinA t - gt^2/2$

⇒ $4.9t^2 - 8.9\times0.8t - 100 = 0$

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m

3 0

2 years ago

3. Our favorite nature explorer Steve Irwin is currenty warking with crocodiles in the Nile River. One of the big boy crocs gets

xeze [42]

Answer:

K.E = 4800 Joules.

Explanation:

Given the following data;

Mass = 600 kg

Velocity = 4 m/s

To find the kinetic energy;

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

$K.E = \frac{1}{2}MV^{2}$

Where;

K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

Substituting into the equation, we have;

$K.E = \frac{1}{2}*600*4^{2}$

$K.E = 300 * 16$

K.E = 4800 Joules.

5 0

2 years ago

What are (a) the lowest frequency, (b) the second lowest frequency, and (c) the third lowest frequency for standing waves on a w

Delvig [45]

(a) The lowest frequency (called fundamental frequency) of a wire stretched under a tension T is given by
$f_1 = \frac{1}{2L} \sqrt{ \frac{T}{m/L} }$
where
L is the wire length
T is the tension
m is the wire mass

In our problem, L=10.9 m, m=55.8 g=0.0558 kg and T=253 N, therefore the fundamental frequency of the wire is
$T= \frac{1}{2L} \sqrt{ \frac{T}{m/L} }= \frac{1}{2 \cdot 10.9 m} \sqrt{ \frac{253 N}{0.0558 kg/10.9 m} }= 10.2 Hz$

b) The frequency of the nth-harmonic for a standing wave in a wire is given by
$f_n = n f_1$
where n is the order of the harmonic and f1 is the fundamental frequency. If we use n=2, we find the second lowest frequency of the wire:
$f_2 = 2 f_1 = 2 \cdot 10.2 Hz=20.4 Hz$

c) Similarly, the third lowest frequency (third harmonic) is given by
$f_3 = 3 f_1 = 3 \cdot 10.2 Hz = 30.6 Hz$

8 0

2 years ago