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3 years ago

11

A roller coaster, traveling with an initial speed of 15 meters per second, decelerates uniformly at â7.0 meters per second2 to a

full stop. approximately how far does the roller coaster travel during its deceleration?

1 answer:

Harrizon [31]3 years ago

8 0

<span>By algebra, d = [(v_f^2) - (v_i^2)]/2a. Thus, d = [(0^2)-(15^2)]/(2*-7) d = [0-(225)]/(-14) d = 225/14 d = 16.0714 m With 2 significant figures in the problem, the car travels 16 meters during deceleration.</span>

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A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s} }{2.8\,s} \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

5 0

3 years ago

On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = go at the north pole

ohaa [14]

The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:
$F=F_g-F_{cf}=mg'-mw^2r'cos(\alpha)\\ ma=mg'-mw^2r'cos(\alpha)\\ a=g'-w^2rcos^2(\alpha)\\$
Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation. $\alpha$ is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.
$g(90)=g_0;\ g_0= g'-w^2rcos^2(90)\\
g_0=g'\\
g(0)=ag_0;\ ag_0=g_0-w^2rcos^2(0)\\
ag_0=g_0-w^2r\\
w^2r=g_0(a-1)
$
Our final equation is:
$g=g_0-g_0(a-1)cos^2(\alpha)$
Colatitude is:
$\alpha_c=90^\circ-\alpha$
The answer is:
$g=g_0-g_0(a-1)cos^2(90-9)=g_0-g_0(a-1)sin^2(9)$

5 0

3 years ago

What is the acceleration of a 59 kg object pushed with a force of 500 Newton's ?

Igoryamba

Answer:

a=500/50=10 m/s^2

Explanation:

f=m•a a=f/m

a=500/50= 10 m/s^2

4 0

2 years ago

A 52 kg and a 95 kg skydiver jump from an airplane at an altitude of 4750 m, both falling in the pike position. Assume all value

Scilla [17]

Answer: 52 kg skydiver: 9.09 m/s and 522.55 s

95 kg skydiver: 12.3 m/s and 386.2 s

Explanation: <u>Drag</u> <u>Force</u> is an opposite force when an object is moving in a fluid.

For skydivers, when falling through the air, the forces acting on it are gravitational and drag forces. At a certain point, drag force equals gravitational force, which is constant on any part of the planet, producing a net force that is zero. Since there is no net force, there is no acceleration and, consequently, velocity is constant. When that happens, the person reached the <u>Terminal</u> <u>Velocity</u>.

Drag Force and Velocity are proportional to the squared speed. So, terminal velocity is given by:

$F_{G}=F_{D}$

$mg=\frac{1}{2}C \rho Av_{T}^{2}$

$v_{T}=\sqrt{\frac{2mg}{\rho CA} }$

where

m is mass in kg

g is acceleration due to gravitational force in m/s²

ρ is density of the fluid in kg/m³

C is drag coefficient

A is area of the object in the fluid in m²

Calculating:

The 52kg skydiver has terminal velocity of:

$v_{T}=\sqrt{\frac{2(52)(9.8)}{(1.21)(0.7)(0.14)} }$

$v_{T}=$ 9.09

The 95kg skydiver's terminal velocity is

$v_{T}=\sqrt{\frac{2(95)(9.8)}{(1.21)(0.7)(0.14)} }$

$v_{T}=$ 12.3

The 52 kg and 95kg skydivers' terminal velocity are 9.09m/s and 12.3m/s, respectively.

The time each one will reach the floor will be:

52 kg at 9.09 m/s:

$t=\frac{4750}{9.09}$

t = 522.5

95 kg at 12.3 m/s:

$t=\frac{4750}{12.3}$

t = 386.2

The 52 kg and 95kg skydivers' time to reach the floor are 522.5 s and 386.2 s, respectively.

3 0

2 years ago

1. write the meaning of the following terms:electrostatic,neutral, positively charged, negatively charged, coulomb,microcoulomb,

dybincka [34]

ELECTROSTATIC:

relating to stationary electric charges or fields as opposed to electric currents.

NEUTRAL:

nor negative nor positive/having no charge

POSITIVELY CHARGED:

positive charge occurs when the number of protons exceeds the number of electrons

NEGATIVELY CHARGED:

negative charge occurs when the number of electrons exceeds the number of protons.

COULOMB:

SI unit for electric charge. One coulomb is equal to the amount of charge from a current of one ampere flowing for one second.

MICROCOULOMB:

a unit of electrical charge equal to one millionth of a coulomb.

NANOCOULOMB:

Nanocoulombs are a unit of charge 1,000,000,000 times smaller than Coulomb.

CONSERVATION OF CHARGE:

constancy of the total electric charge in the universe or in any specific chemical or nuclear reaction

QUANTISATION OF CHARGE:

Charge quantization is the principle that the charge of any object is an integer multiple of the elementary charge.

5 0

2 years ago