If HST has a tangential speed of 7,750 m/s, how long is HST’s orbital period? The radius of Earth is 6.38 × 106 m. s

A neutron star that spins and emits beams of energy is called a _____.

salantis [7]

Pulsar or also sometimes as a actively rotating accretion disc

5 0

3 years ago

An unstrained horizontal spring has a length of 0.39 m and a spring constant of 350 N/m. Two small charged objects are attached

Vera_Pavlovna [14]

Answer:

A) The possible algebraic signs will either be both positive (+) or both negative (-) charged since the 2 objects are repelling each other to stretch the string.

B) Magnitude of charges = 1.206 × 10^(-6) C

Explanation:

We are given;

Spring constant;k = 350 N/m

Spring length;L = 0.39 m

Stretched length of spring;x = 0.022 m

A) The spring stretches by 0.022m. Therefore, the total force is (350 × 0.022) N = 7.7N. The charged objects will either be both positive (+) or both negative (-) charged since they are repelling each other to stretch the string.

B) Force (F) required to stretch spring is given by the formula;

F = kx

Thus:

F = (350 × 0.022)

F = 7.7 N

Now, if we assume point charges, then the distance (r) between them will be given as:

r = (0.39 + 0.022) = 0.412 m

Coulomb's Law has a formula:

F = k(q1×q2)/r²

where k is coulomb's constant = 8.99 × 10^(9) Nm²/C²

Making q1 × q2 the subject, we have;

(q1 × q2) = Fr²/k = 7.7 × 0.412²/(8.99 × 10^(9))

(q1 × q2) = 14.54 × 10^(-11) C

We are told that both charges are equal, thus; |q1| = |q2|

So;

q = √(14.54 × 10^(-11)) = 1.206 × 10^(-6) C

6 0

3 years ago

Two gliders collide on a frictionless air track that is aligned along the x axis. Glider A has an initial velocity of +4.0 m/s a

DedPeter [7]

Answer:

As collision is elastic,thus we can use conservation of momentum equation

mA=0.2 kg

(vB)1=0 m/s.......................as it is on rest before collision

(vA)1=4 m/s

(vA)2=-1 m/s

(vB)2=2 m/s

using equation

(mA*vA+mB*vB)1= (mA*vA+mB*vB)2

Where 1 and 2 represents before and after collision

(0.2*4)+(mB*0)=(0.2*-1)+(mB*2)

0.8=-0.2+(2mB)

mass of object B=mB=0.3 Kg

6 0

3 years ago

To what potential should you charge a 3.0 μf capacitor to store 1.0 j of energy?

Nimfa-mama [501]

The energy stored in a capacitor is given by:
$U= \frac{1}{2}CV^2$
where
U is the energy
C is the capacitance
V is the potential difference

The capacitor in this problem has capacitance
$C=3.0 \mu F = 3.0 \cdot 10^{-6} F$
So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
$V= \sqrt{ \frac{2U}{C} }= \sqrt{ \frac{2 \cdot 1.0 J}{3.0 \cdot 10^{-6}F} }=816 V$

8 0

3 years ago

Calculate the current flowing if a charge of 36 kilocoulombs flows in 1 hour.

o-na [289]

Answer:

Current = 10 Amperes.

Explanation:

Given the following dat;

Quantity of charge, Q = 36 kilocoulombs (KC) = 36 * 1000 = 36000C

Time = 1 hour to seconds = 60*60 = 3600 seconds

To find the current;

Quantity of charge = current * time

Substituting in the equation

36000 = current * 3600

Current = 36000/3600

Current = 10 Amperes.

6 0

3 years ago