What force is required to accelerate a 1,000 kg car 5 m/s2?

A Hooke's law spring is compressed 12.0 cm from equilibrium, and the potential energy stored is 72.0 J. What compression (as mea

Anuta_ua [19.1K]

Answer:14.14 cm

Explanation:

Given

Spring Compression $x=12 cm$

Potential energy Stored in spring $=72 J$

Suppose k is the spring constant of spring

Potential Energy of spring is given by $=\frac{kx^2}{2}$

$\frac{k(0.12)^2}{2}=72$

$k(0.12)^2=144$

$k=10,000 N/m$

$k=10 kN/m$

for 100 J energy

$\frac{k(x_0)^2}{2}=100$

$10\times 10^3\cdot (x_0)^2=200$

$(x_0)^2=2\times 10^{-2}$

$x_0=0.1414$

$x_0=14.14 cm$

6 0

2 years ago

A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.480 m/s. The to

SpyIntel [72]

This is a problem of conservation of momentum

Momentum before throwing the rock: m*V = 96.0 kg * 0.480 m/s = 46.08 N*s

A) man throws the rock forward

=>

rock:
m1 = 0.310 kg
V1 = 14.5 m/s, in the same direction of the sled with the man

sled and man:
m2 = 96 kg - 0.310 kg = 95.69 kg
v2 = ?

Conservation of momentum:
momentum before throw = momentum after throw

46.08N*s = 0.310kg*14.5m/s + 95.69kg*v2

=> v2 = [46.08 N*s - 0.310*14.5N*s ] / 95.69 kg = 0.434 m/s

B) man throws the rock backward

this changes the sign of the velocity, v2 = -14.5 m/s

46.08N*s = - 0.310kg*14.5m/s + 95.69kg*v2

v2 = [46.08 N*s + 0.310*14.5 N*s] / 95.69 k = 0.529 m/s

3 0

2 years ago

I need help with this question how to solve it for Brass and Cooper

Ksenya-84 [330]

Take into account that density and relative density are given by:

$\begin{gathered} \text{density}=\text{ mass/volume} \\ \text{relative density = density/density of water} \end{gathered}$

Take into account that the volume associated to each of the given sustances in the table is determined by the Level Difference (because it is the change in the volume of the water of the recipient in which the substance is immersed).

The density of water in kg/m^3 is 1000 kg/m^3.

Due to the density must be given in kg/m^3, it is necessary to express the volumes of the table in m^3 and mass in kg, then, consider the following conversion factor:

1 m^3 = 1000000 ml

1 kg = 1000 g

Then, you obtain the following results:

Brass:

$\begin{gathered} 53.2g\cdot\frac{1kg}{1000g}=0.0532kg \\ 6ml\cdot\frac{1m^3}{1000000ml}=0.000006m^3 \\ \text{density}=\frac{0.0532kg}{0.000006m^3}\approx8866.67\frac{kg}{m^3} \\ \text{relative density=}\frac{(\frac{8866.66kg}{m^3})}{(1000\frac{kg}{m^3})}\approx8.87 \end{gathered}$

Cooper:

$\begin{gathered} 57.4g=0.0574kg \\ 6ml=0.000006m^3 \\ \text{density}=\frac{0.0574kg}{0.000006m^3}\approx9566.67\frac{kg}{m^3} \\ \text{relative density=}\frac{\frac{9566.67kg}{m^3}}{1000kg}=9.57 \end{gathered}$

3 0

6 months ago

6) A map in a ship’s log gives directions to the location of a buried treasure. The starting location is an old oak tree. Accord

kiruha [24]

Answer:

Sorry cant find the answer but i hope you got it right and if you didn't you'll still do great. :)

Explanation:

4 0

2 years ago

Para que el calor sea transmitido por conducción los cuerpos deben tocarse?<br> verdadero o falso

dybincka [34]

Answer:

verdadero

Explanation:

Conducción:Es la más sencilla de entender, consiste en la transferencia de calor entre dos puntos de un cuerpo que se encuentran a diferente temperatura, sin que se produzca transferencia de materia entre ellos.

7 0

2 years ago