A man walks 6 hrs from his house to his office at an average speed of 2km/hr.find the difference between his house and office

A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh

alekssr [168]

(a) 392 N/m

Hook's law states that:

$F=k\Delta x$ (1)

where

F is the force exerted on the spring

k is the spring constant

$\Delta x$ is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

$F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N$

- The stretching of the spring is

$\Delta x=15 cm-10 cm=5 cm=0.05 m$

Solving eq.(1) for k, we find the spring constant:

$k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m$

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

$F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N$

And so, the stretch of the spring is

$\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm$

And since the initial lenght of the spring is

$x_0 = 10 cm$

The final length will be

$x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm$

8 0

3 years ago

Read 2 more answers

What is the pressure exerted by a column of fresh water 10m high (p=1000kg/m3 and g=10m/s)

alexira [117]

Height (h) = 10 m
Density (ρ) = 1000 Kg/m^3
Acceleration due to gravity (g) = 10 m/s^2
We know, pressure in a fluid = hρg
Therefore, the pressure exerted by a column of fresh water
= hρg
= (10 × 1000 × 10) Pa
= 100000 Pa

Answer:

100000 Pa

Hope you could understand.

If you have any query, feel free to ask.

8 0

2 years ago

A 14.0 gauge copper wire of diameter 1.628 mm carries a current of 12.0 mA . A) What is the potential difference across a 1.80 m

Serga [27]

Answer: a) 139.4 μV; b) 129.6 μV

Explanation: In order to solve this problem we have to use the Ohm law given by:

V=R*I whre R= ρ *L/A where ρ;L and A are the resistivity, length and cross section of teh wire.

Then we have:

for cooper R=1.71 *10^-8* 1.8/(0.001628)^2= 11.61 * 10^-3Ω

and for silver R= 1.58 *10^-8* 1.8/(0.001628)^2=10.80 * 10^-3Ω

Finalle we calculate the potential difference (V) for both wires:

Vcooper=11.62* 10^-3* 12 * 10^-3=139.410^-6 V

V silver= 10.80 10^-3* 12 * 10^-3=129.6 10^-6 V

8 0

3 years ago

Read 2 more answers

Which of the following can reduce biodiversity within a natural environment? A. Yucca moths pollinate and feed on yucca plants i

DaniilM [7]

I wanna say B Red-billed oxpeckers eat ticks off of impalas

8 0

3 years ago

The magnetic field at the center of a 1.50-cm-diameter loop is 2.70 mT . Part A. What is the current in the loop?

elixir [45]

Explanation:

It is given that,

Diameter of the circular loop, d = 1.5 cm

Radius of the circular loop, r = 0.0075 m

Magnetic field, $B=2.7\ mT=2.7\times 10^{-3}\ T$

(A) We need to find the current in the loop. The magnetic field in a circular loop is given by :

$B=\dfrac{\mu_o I}{2r}$

$I=\dfrac{2Br}{\mu_o}$

$I=\dfrac{2\times 2.7\times 10^{-3}\times 0.0075}{4\pi \times 10^{-7}}$

I = 32.22 A

(b) The magnetic field on a current carrying wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$r=\dfrac{\mu_o I}{2\pi B}$

$r=\dfrac{4\pi \times 10^{-7}\times 32.22}{2\pi \times 2.7\times 10^{-3}}$

r = 0.00238 m

$r=2.38\times 10^{-3}\ m$

Hence, this is the required solution.

8 0

3 years ago