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3 years ago

9

Please......... I need desperate help

1 answer:

PIT_PIT [208]3 years ago

3 0

' A ' is one crest of the wave. After every wavelength, there's another one.

' B ' . . . the vertical arrow under B shows the amplitude of the wave

' C ' is one trough of the wave. After every wavelength, there's another one.

' D ' . . . the horizontal arrow over D shows the wavelength.

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Physics!!! please help D:

Ugo [173]

Answer: A

Explanation: isotopes of the same thing element have the same number of protons in the nucleus but differ in the number of neutrons.

6 0

2 years ago

Two scientists interpret the same set of data differently. How would the scientific community deal with this problem?

alexdok [17]

The correct answer would be D. A new experiment would be needed to be done in order to test the conclusions. In science there is no authority, data is the only thing that matters. So if we have two different conclusions from the same date the only solution is to perform more tests and more experiments to see what is correct.

7 0

3 years ago

g A box having mass 0,5kg is placed in front of a 20 cm compressed spring. When the spring released, box having mass m1, collide

vagabundo [1.1K]

Answer:Expression given below

Explanation:

Given mass of spring $\left ( m_1\right )=0.5 kg$

Compression in the spring $\left ( x\right )=20 cm$

Let the spring constant be K

Using Energy conservation

potential energy stored in spring =Kinetic energy of Block $\left ( m_1\right )$

$\frac{1}{2}Kx^2=\frac{1}{2}m_1v^2$

$v=x\sqrt{\frac{k}{m_1}}$

now conserving momentum

$m_1v=\left ( m_1+m_2\right )v_0$

$v_0=\frac{m_1}{m_1+m_2}v$

where $v_0$ is the final velocity

3 0

3 years ago

Two identical small charged spheres hang in equilibrium with equal masses (0.02kg). The length of the strings is equal (0.18m) a

Yuliya22 [10]

Answer:

The value is $q = 3.4 *10^{-6} \ C$

Explanation:

From the question we are told that

The mass of each sphere is $m_1 = m_2 = m = 0.020 \ kg$

The length of the string is $l = 0.18 \ m$

The angle of with the vertical is $\theta = 7^o$

The acceleration due to gravity is $g = 9.8 \ m/s^2$

Generally the force acting between the forces is mathematically represented as

$F = T cos \theta = \frac{k* q^2}{ r^2}$

=> $T cos \theta = \frac{k* q^2}{ r^2}$

Generally from Pythagoras theorem the radius of the circular curve created by the force is

$r = 2 L sin (\theta )$

=> $r = 2* 0.180 sin (7)$

=> $r = 0.043 \ m$

=> $q = tan \theta * \frac{m * g * r^2 }{k}$

=> $q = tan(7)* \frac{ 0.02 * 9.8 * 0.043^2 }{9*10^{9}}$

=> $q = 3.4 *10^{-6} \ C$

7 0

3 years ago

a shell fired from a cannon at 60 ° from horizontal strikes a target 20m high at a distance 80m. Calculate the initial velocity

stich3 [128]

consider the motion along the X-direction

X = horizontal displacement = 80 m

$V_{ox}$ = initial velocity along the x-direction = v Cos60

t = time of travel

using the equation

X = $V_{ox}$ t

80 = (v Cos60) (t)

t = 160/v eq-1

consider the motion in vertical direction :

Y = vertical displacement = 20 m

$V_{oy}$ = initial velocity in Y-direction = v Sin60

a = acceleration = - 9.8 m/s²

t = time of travel = 160/v

using the equation

Y = $V_{oy}$ t + (0.5) a t²

20 = (v Sin60) (160/v) + (0.5) (- 9.8) (160/v)²

v = 32.5 m/s

4 0

3 years ago