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Alecsey [184]
3 years ago
8

A person with a mass of 72 kg is riding a bicycle is accelerating at a rate of 5m/s on a horizontal surface. What is the weight

force the person exerts on the bicycle seat
Physics
1 answer:
nikklg [1K]3 years ago
7 0

Answer:

w = 706.32 [N]

Explanation:

The force due to gravitational acceleration can be calculated by means of the product of mass by gravitational acceleration.

w = m*g

where:

w = weight [N] (units of Newtons)

m = mass = 72 [kg]

g = gravity acceleration = 9.81 [m/s²]

Then we have:

w = 72*9.81\\w = 706.32 [N]

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Two pieces of evidence for plate tectonics?
Reika [66]
Evidence from fossils, glaciers, and complementary coastlines helps reveal how the plates once fit together. Fossils tell us when and where plants and animals once existed. Some life "rode" on diverging plates, became isolated, and evolved into new species.
5 0
3 years ago
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Polar molecules have ionic bonds.<br> True<br> False
Nadusha1986 [10]
Polar molecules do have ionic bonds
6 0
3 years ago
The voltage between two parallel plates separated by a distance of 3. 0 cm is 120 v. The electric field between the plates is?
Lapatulllka [165]

The electric field between plates is 4000V/m.

An electric field (sometimes E-field) is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It also refers to the physical field for a system of charged particles.

The value of the electric field has dimensions of force per unit charge. In the metre-kilogram-second and SI systems, the appropriate units are newtons per coulomb, equivalent to volts per metre.

The voltage between points A and B is

V=E.d

E =V/d  (uniform E- field only)  

where  d is the distance from A to B, or the distance between the plates.

Given:

distance d = 3 cm

voltage V = 120 V

Electric field E = V/d

                     E = 120 V / 3cm

                     E = 40 V / 1 cm               [ 1 cm = 1/100 m ]

                    E = 4000 V/m.

Learn more about Electric field here:

brainly.com/question/8971780

#SPJ4

6 0
1 year ago
A bottle lying on the windowsill falls off and takes 4.95 seconds to reach the ground. The distance from the windowsill to the g
mr Goodwill [35]

The distance an object falls from rest through gravity is

                         D  =  (1/2) (g) (t²)

            Distance  =  (1/2 acceleration of gravity) x (square of the falling time)

We want to see how the time will be affected
if  ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '.

                                                      D  =  (1/2) (g) (t²)

Multiply each side by  2 :         2 D  =            g    t² 

Divide each side by ' g ' :      2 D/g =                  t²

Square root each side:        t = √ (2D/g)


Looking at the equation now, we can see what happens
to ' t ' when only ' g ' changes: 

-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'
                                             and smaller 'g' ==> longer 't' .

-- They don't change by the same factor, because  1/g  is inside
the square root.  So 't' changes the same amount as  √1/g  does.

Gravity on the surface of the moon is roughly  1/6  the value
of gravity on the surface of the Earth.

So we expect ' t ' to increase by  √6  =  2.45 times.

It would take the same bottle  (2.45 x 4.95) = 12.12 seconds
to roll off the same window sill and fall 120 meters down to the
surface of the Moon.
4 0
3 years ago
Read 2 more answers
*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude
kvv77 [185]

Answer:

Approximately 3.86\; {\rm N} (given that the magnitude of this charge is -7.35\; {\rm \mu C}.)

Explanation:

If a charge of magnitude q is placed in an electric field of magnitude E, the magnitude of the electrostatic force on that charge would be F = E\, q.

The magnitude of this charge is q = 7.35\; {\rm \mu C}. Apply the unit conversion 1\; {\rm \mu C} = 10^{-6}\; {\rm C}:

\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}.

An electric field of magnitude E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}} would exert on this charge a force with a magnitude of:

\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}.

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0
2 years ago
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