All categories

3 years ago

Calculate the force it would take to accelerate a 250 kg motorcycle at a rate of 5 m/s2

Physics

2 answers:

Sphinxa [80]3 years ago

7 0

Force = mass x acceleration

Mass is given to be 250kg

Acceleration is given to be 5m/s^2

Force = 250kg x 5m/s^2 = 1,250 kgm/s^2 = 1,250 Newtons

evablogger [386]3 years ago

5 0

Answer:

1,250

Explanation:

You might be interested in

A body with a mass of 2,000 kg is lifted to a height of 15 m within a time of 15 s. Which one of the following statements concer

Aleksandr [31]

<span>So we want to know which statement is true for the body of mass m=2000kg that is lifted to a height of h=15m in t=15 s. Lets calculate each of the following: Gravity force on the body is F=m*g=2000*9.81=19620 N so a is FALSE. Potential energy of the body when it is lifted to the height of 15 m is Ep=m*g*h=2000*9.81*15=294300 J so b is FALSE. Work to lift the body is: W=Fg*h=2000*9.81*15= Ep=294300 J so c is FALSE. Power P=W/t=294300/15=19620 W So d is TRUE. </span>

8 0

2 years ago

Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.7 A. The resistanc

lisov135 [29]

Answer:

a) The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b)The mass per unit length \lambdaλ for a copper cable is 0.757kg/m

c)The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d)The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

Explanation:

The expression for electric field of conductor is,

$E = \frac{V}{L}$

The general equation of voltage is,

V = iR

The expression for current density in term of electric field is,

$J = \frac{E}{p}$

Substitute (V/L) for E in the above equation of current density.

$J = \frac{V}{pL} ------(1)$

Substitute iR for V in equation (1)

$J = \frac{iR}{pL} ------(2)$

Substitute 1.69 × 10⁸ Ω .m for p

50A for i

0.200Ω.km⁻¹ for (R/L) in eqn (2)

$J = \frac{(50) (0.200\times 10^-^3) }{1.69 \times 10^-^8 } \\\\= 5.91 \times 10^5A.m^-^2$

The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b) The expression for resistivity of the conductor is,

$p = \frac{RA}{L}$

$A = \frac{pL}{R}$

The expression for mass density of copper is,

m = dV

where, V is the density of the copper.

Substitute AL for V in equation of the mass density of copper.

m=d(AL)

m/L = dA

λ is use for (m/L)

substitute,

pL/R for A and λ is use for (m/L) in the eqn above

$\lambda = d\frac{p}{\frac{R}{L} } ------(3)$

Substitute 0.200Ω.km⁻¹ for (R/L)

8960kgm⁻³ for d and 1.69 × 10⁸ Ω .m

$\lambda = (8960) \frac{(1.69 \times 10^-^8 }{0.200\times 10^-^3} \\\\= 0.757kg.m^-^1$

c) Using the equation (2) current density for aluminum cable is,

$J = \frac{iR}{pL}$

p is the resistivity of the aluminum cable.

Substitute 2.82 × 10⁻⁸Ω.m for p ,

50A for i and 0.200Ω.km⁻¹ for (R/L)

$J = \frac{(50)(0.200\times10^-^3) }{2.89\times 10^-^8} \\\\= 3.5 \times10^5A/m^2$

The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d) Using the equation (3) mass per unit length for aluminum cable is,

$\lambda = d\frac{p}{\frac{R}{L} }$

p is the resistivity and is the density of the aluminum cable.

Substitute 0.200Ω.km⁻¹ for (R/L), 2700 for d and 2.82 × 10⁻⁸Ω.m for p

$\lambda = (2700) \frac{(2.82 \times 10^-^8) }{(0.200 \times 10^-^3) } \\\\= 0.380kg/m$

The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

7 0

3 years ago

Read 2 more answers

Kalea throws a baseball directly upward at time t = 0 at an initial speed of 13.7 m/s. How high h does the ball rise above its r

Trava [24]

Answer:

h = 9.57 seconds

Explanation:

It is given that,

Initial speed of Kalea, u = 13.7 m/s

At maximum height, v = 0

Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :

$v=u-gt$

$u=gt$

$t=\dfrac{u}{g}$

$t=\dfrac{13.7}{9.8}$

t = 1.39 s

Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :

$h=ut+\dfrac{1}{2}at^2$

Here, a = -g

$h=ut-\dfrac{1}{2}gt^2$

$h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2$

h = 9.57 meters

So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.

7 0

3 years ago

2. Which of the following accurately describes the effects of temperature changes on solids? A. If the temperature of a solid is

balandron [24]

<span>B. If the temperature of a solid is increased, the solid expands and its volume increases.</span>

3 0

3 years ago

Read 2 more answers

When looking through a fish tank, the mediums which light passes through are:

Alinara [238K]

The answer is : C ) air,water,and the tank glass.
-Hope this helps.

3 0

3 years ago

Read 2 more answers