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Deffense [45]
3 years ago
7

Give you point but please i need help in physcis

Physics
1 answer:
Anastasy [175]3 years ago
7 0

Explanation:

Hi

Hi

Hi

Hi

BRAINILIEST PLEASE

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Does a comets tail always trail along behind it in its orbit?
Marat540 [252]
No, it only does when entering an atmosphere
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3 years ago
A bicyclist of mass 112 kg rides in a circle at a speed of 8.9 m/s. If the radius of the circle is 15.5 m, what is the centripet
kogti [31]
The centripetal force, Fc, is calculated through the equation, 
                                    Fc = mv²/r
where m is the mass,v is the velocity, and r is the radius. 
Substituting the known values,
                                     Fc = (112 kg)(8.9 m/s)² / (15.5 m)
                                         = 572.36 N
Therefore, the centripetal force of the bicyclist is approximately 572.36 N. 
4 0
3 years ago
Read 2 more answers
A parallel plate capacitor creates a uniform electric field of and its plates are separated by . A proton is placed at rest next
zalisa [80]

Complete Question

A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?

Answer:

V=1.4*10^5m/s

Explanation:

From the question we are told that:

Electric field B=1.5*10N/C

Distance d=2 x 10^{-3}

At negative plate

Generally the equation for Velocity is mathematically given by

V^2=2as

Therefore

V^2=\frac{2*e_0E*d}{m}

V^2=\frac{2*1.6*10^{-19}(5*10^4)*2 * 10^{-3}}{1.67*10^{-28}}

V=\sqrt{19.2*10^9}

V=1.4*10^5m/s

5 0
3 years ago
1. Applied research observational evidence 2. Basic research the experimental factor that changes in response to a change in the
lesya [120]

Answer:

This question is about matching each definition with its correct term. Please find the term matched with their appropriate definition below.

Explanation:

1. Empirical evidence: An empirical evidence is an observational evidence i.e an evidence gathered by observation or use of senses.

2. Dependent variable: Dependent variable is an experimental factor that changes in response to a change in the independent variable. In other words, it is dependent on the independent variable.

3. Applied research: Applied research is a type of research oriented at solving a present problem or need. It encompasses the production of products that can be sold for profit.

4. Hypothesis: A hypothesis in an experiment is a proposed explanation for a scientific problem that itself can be tested by experimentation. A hypothesis aims at providing a testable explanation to an observed problem.

5. Control: A control is a quantity in an experiment that remains unchanged or constant. It is kept the same by the experimenter for all groups in the experiment in order not to influence the outcome.

6. Basic research: Basic research is the research that expands knowledge in a particular area. It is the kind of research that aims at filling a knowledge void or satiating curiosity.

7. Independent variable: The independent variable is the experimental factor that is changed or manipulated deliberately by the scientist.

8 0
3 years ago
A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
2 years ago
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