All categories

3 years ago

How much work is done by a forklift raising a 125 kg object a distance of 10 meters?

Physics

1 answer:

lozanna [386]3 years ago

6 0

W = Fd

W = 1225 N x 10 m = 12250

You might be interested in

What is the momentum of a 10kg ball moving at a velocity of 4m/s east?

Katyanochek1 [597]

Answer:

40 kg.m/s

Explanation:

Momentum, p is defined as the product of mass and velocity of an object. Numerically, it is represented as, p=mv where m is mass of the object and v is the velocity in which the object moves, with keen observation on the direction before and after collision. Substituting 10 kg for m and 4 m/s for v then momentum, P=10*4=40 kg.m/s

6 0

3 years ago

What is Newtons third law

Vilka [71]

Newton's third law<span> is for every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. </span>

3 0

3 years ago

Read 2 more answers

An aluminum cup contains 225 g of water and a 40-g copper stirrer, all at 27Â°C. A 470-g sample of silver at an initial temperat

vagabundo [1.1K]

Answer:

Mal = 0.232 kg = 232 g

Explanation:

mass of water (Mw) = 225 g = 0.225 kg

mass of copper stirrer (Mcu) = 40 g = 0.04 kg

initial temperature of water (Tw) = 27 degrees

mass of silver (Mag) = 470 g = 0.47 kg

initial temperature of silver (Ts) = 85 degrees

final temperature of the mixture (T) = 32 degrees

find the mass of the aluminum cup (Mal)

applying the conservation of energy

((Mal.cAl) + (Mw.cW) + (Mcu.cCu))(ΔTw) = (Mag.cAg)(ΔTag)

we require the specific heat capacities of water (cW) , aluminium (cAl) , copper (cCu) and silver (cAg) which are as follows

water (cW) =4186 J/kg.K

aluminium (cAl) = 900 J/kg.K

copper (cCu) = 387 J/kg.K

silver (cAg) = 234 J/kg.K

now we can put in our values to get the mass of the Aluminium cup (Mag)

((Mal.900) + (0.225 x 4186) + (0.04 x 387))(32-27) = (0.47 x 234)(85-32)

(900 . Mal + 957.33) x 5 = 5828.94

900 .Mal + 957.33 = 1165.79

900.Mal = 1165.79 - 957.33 = 208.5

Mal = 0.232 kg = 232 g

8 0

3 years ago

Spacetime interval: What is the interval between two events if in some given inertial reference frame the events are separated b

shepuryov [24]

Answer:

a. $\Delta s ^2 = 8.0888 \ 10^{17} m^2$

b. $\Delta s ^2 = 3.0234 \ 10^{16} m^2$

c. $\Delta s ^2 = 3.0234 \ 10^{20} m^2$

Explanation:

The spacetime interval $\Delta s^2$ is given by

$\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2$

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

$\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2$ .

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

$\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 5,625 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2$

$\Delta (c t) ^ 2 = (899,377,374 \ m)^2$

$\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2$

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2$

$\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2$

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2$

$\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2$

5 0

3 years ago

A 460 g , 6.0-cm-diameter can is filled with uniform, dense food. It rolls across the floor at 1.1 m/s . Part A What is the can'

Reika [66]

Answer:

the can's kinetic energy is 0.42 J

Explanation:

given information:

Mass, m = 460 g = 0.46 kg

diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m

velocity, v = 1.1 m/s

the kinetic energy of the can is the total of kinetic energy of the translation and rotational.

KE = $\frac{1}{2}$ I ω^2 + $\frac{1}{2} mv^{2}$

where

I = $\frac{1}{2} mr^{2}$ and ω = $\frac{v}{r}$

thus,

KE = $\frac{1}{2}$ $\frac{1}{2} mr^{2}$ ( $\frac{v}{r}$ )^2 + $\frac{1}{2} mv^{2}$

= $\frac{1}{2}$ $\frac{1}{2} mr^{2}$ $\frac{v^{2} }{r^{2}}$ + $\frac{1}{2} mv^{2}$

= $\frac{1}{4} mv^{2}$ + $\frac{1}{2} mv^{2}$

= $\frac{3}{4} mv^{2}$

= $\frac{3}{4} (0.46) (1.1)^{2}$

= 0.42 J

8 0

3 years ago