All categories

3 years ago

Help with the following five problem please! For number 12 the teacher said 10 drops= 1 mm

Physics

1 answer:

Whitepunk [10]3 years ago

7 0

a 100mg dog ???

a 0.1 gram dogf ???

hmmm no.

You might be interested in

A car accelerates at 4 m/s/s from rest. What is the car's velocity after it travels 20 m?

MrRa [10]

The cars velocity after it travels 20 m is 4 because you have to divide it by the miles that the car is traveling and the 4 miles from the rest; and your finally answer being 5
Answer: 5

3 0

2 years ago

Taking the resistivity of platinoid as 3.3 x 10-7 m, find the resistance of 7.0 m of platinoid wire of average diameter 0.14 cm.

mr_godi [17]

Answer:

$1.5 \Omega$

Explanation:

The resistance of a wire is given by the equation:

$R=\rho \frac{L}{A}$

where

$\rho$ is the resistivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

In this problem, we have a wire of platinoid, whose resistivity is

$\rho = 3.3\cdot 10^{-7} \Omega m$

The length of the wire is

L = 7.0 m

And its radius is

$r=\frac{0.14 cm}{2}=0.07 cm = 7\cdot 10^{-4} m$ , so the cross-sectional area is

$A=\pi r^2=\pi(7\cdot 10^{-4})^2=1.54\cdot 10^{-6}m^2$

Solving for R, we find the resistance of the wire:

$R=(3.3\cdot 10^{-7})\frac{7.0}{1.54\cdot 10^{-6}}=1.5 \Omega$

3 0

2 years ago

A rocket accelerates straight up from the ground at 12.6 m/s^2 for 11.0 s. Then the engine cuts off and the rocket enters free f

FrozenT [24]

Answer:

a) 138.6 m/s

b) 762.3 m

c) 122.3 m/s

d) 24.47

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v=u+at\\\Rightarrow v=0+12.6\times 11\\\Rightarrow v=138.6 \ m/s$

Velocity at the end of its upward acceleration is 138.6 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 12.6\times 11^2\\\Rightarrow s=762.3\ m$

Maximum height the rocket reaches is 762.3 m

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 762.3-0^2}\\\Rightarrow v=122.3\ m/s$

The velocity with which the rocket crashes to the Earth is 122.3 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow 762.3=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{762.3\times 2}{9.81}}\\\Rightarrow t=12.47\ s$