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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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A 50.0 kg driver is riding at 35.0 m/s in her red sports car when she must suddenly slam on the brakes to avoid
egoroff_w [7]

Answer:

3500N

Explanation:

Given parameters:

Mass of driver  = 50kg

Speed  = 35m/s

Time  = 0.5s

Unknown:

Average force the seat belt exerts on her = ?

Solution:

The average force the seat belt exerts on her can be deduced from Newton's second law of motion.

   F = mass x acceleration

So;

     F  = mass x  \frac{change in velocity }{time}

  F  = 50 x \frac{35}{0.5}    = 3500N

8 0
3 years ago
An object is five focal lengths from a concave mirror.how do the object and image heights compare?
enot [183]

An object distance is presented as s = 5f and we know that the mirror equation relates the image distance to the object distance and the focal length.

The mirror equation is 1/f = 1/s + 1/s’ where the variable f stands for the focal length of the mirror. Variable (s) represents the distance between the mirror surface and the object and the variable <span>(s’) represents the distance between the mirror surface and the image. </span>

In addition, a concave mirror will have a positive focal length (f) and a convex mirror will have a negative focal length (f).

Now, we then have 1/f = 1/5f + 1/s’ which is s’ = 5f/4

Then we get the magnification ratio that expresses the size or amount of magnification or reduction of the object or image and to get the magnification, we use this equation: M= s’/s

M= 5f/4x5f

s’ = 1/4s

Therefore, the image height is one fourth of the object height

7 0
3 years ago
1. On each of your equipotential maps, draw some electric field lines with arrow heads indicating the direction of the field. (H
JulijaS [17]

Answer:

The angle between the electric field lines and the equipotential surface is 90 degree.

Explanation:

The equipotential surfaces are the surface on which the electric potential is same. The work done in moving a charge from one point to another on an equipotential surface is always zero.

The electric field lines are always perpendicular to the equipotential surface.

As

dV = \overrightarrow{E} . d\overrightarrow{r}\\\\

For equipotential surface, dV = 0 so

0 = \overrightarrow{E} . d\overrightarrow{r}\\\\

The dot product of two non zero vectors is zero, if they are perpendicular to each other.

5 0
2 years ago
Did I do these questions correctly?
SOVA2 [1]
Yes, they seem right to me.
4 0
3 years ago
Calculate the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lig
Nana76 [90]

Answer:

Work done = 35467.278 J

Explanation:

Given:

Height of the cone = 4m

radius (r) of the cone = 1.2m

Density of the cone = 600kg/m³

Acceleration due to gravity, g = 9.8 m/s²

Now,

The total mass of the cone (m) = Density of the cone × volume of the cone

Volume of the cone = \frac{1}{3}\pi r^2 h

thus,

volume of the cone = \frac{1}{3}\pi 1.2^2\times 4 = 6.03 m³

therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg

The center of mass for the cone lies at the \frac{1}{4}times the total height

thus,

center of mass lies at,  h' = \frac{1}{4}\times4=1m

Now, the work gone (W) against gravity is given as:

W = mgh'

W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J

4 0
3 years ago
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