Please help thanks

Which of these experiments would make use of qualitative data

7nadin3 [17]

It has to due with numbers so I would say the last one!

5 0

3 years ago

Read 2 more answers

I need help someone help me

GaryK [48]

what do you need , i mean your help , let me see if i can help

5 0

3 years ago

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

The mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

The distance apart is $d = 0.4 \ m$

The speed of the second sphere is $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.8 \ m$ is mathematically represented

$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

$KE = \frac{1 }{2} m (v_1)^2$

substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

$KE = 0.3 \ J$

And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

$U = 0.18 \ J$

So

$Q = 0.3 + 0.18$

$Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

$KE_f = \frac{1 }{2} m (v_2^2$

substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

$KE_f = 7.50 *10^{ -4} (v_2 )^2$

And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

$U_f = 0.36 \ J$

From the law of energy conservation

$Q = Q_f$

So

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

6 0

3 years ago

A radar antenna is tracking a satellite orbiting the earth. At a certain time, the radar screen shows the satellite to be 118 km

ruslelena [56]

Answer:

x component 60.85 m

y component 101.031 m

Explanation:

We have given distance r = 118 km

Angle which makes from ground = 58.9°

(a) X component of distance is given by $r_x=rcos\Theta =118\times cos58.9=118\times 0.5165=118=60.85m$

(b) Y component of distance is given by $r_Y=rcos\Theta =118\times sin58.9=118\times 0.8562=101.0316m$

These are the x and y component of position vector

6 0

3 years ago

Please help me feel in the blanks Giving 30 points

maw [93]

4:chemical properties can only be observed when a substance changes into another substance.

5: physical properties such as color and shape are easy to observe

6: in a chemical change an altered substance forms

7: cooking or baking food will result in a chemical change

8: a melting ice cube is a physical change

9: the rusting of iron is a chemical change

10: water boiling is a scientific physical change

7 0

3 years ago