Ask question

Ask question

All categories

3 years ago

9

You have landed on an unknown planet, Newtonia, and want to know what objects will weigh there. You find that when a certain too

l is pushed on a frictionless horizontal surface by a 12.0 N force, it moves 16.4 m in the first 2.50 s, starting from rest. You next observe that if you release this tool from rest at 10.3 m above the ground, it takes 2.88 s to reach the ground.What does the tool weigh on Newtonia (in N)?What would it weigh on Earth (in N)?

2 answers:

steposvetlana [31]3 years ago

8 0

Answer:

$w'=5.679\ N$ on the planet

$w=22.43\ N$ on earth

Explanation:

Given:

initial velocity of the tool before pushing, $u=0\ m.s^{-1}$

force applied on the tool, $F=12\ N$

displacement of the tool, $s=16.4\ m$

time taken for the displacement, $t=2.5\ s$

height of releasing the tool, $h=10.3\ m$

time taken by the tool to fall on the ground, $t_v=2.88\ s$

<u>Now using the equation of motion:</u>

$s=u.t+\frac{1}{2}a.t^2$

where:

a = acceleration of the object

$16.4=0+0.5\times a\times 2.5^2$

$a=5.248\ m.s^{-2}$

Now the mass of the tool:

$m=\frac{F}{a}$

$m=\frac{12}{5.248}$

$m=2.2866\ kg$

<u>Using the equation of motion when the tool is dropped:</u>

$h=u.t_v+\frac{1}{2} \times g.t_v^2$

here:

g = acceleration due to gravity on the planet

$10.3=0+0.5\times a\times 2.88^2$

$g=2.4836\ m.s^{-2}$

Weight of the tool in the planet:

$w'=m.g$

$w'=2.2866\times 2.4836$

$w'=5.679\ N$

Weight of the tool on the earth:

$w=m.g'$

$w=2.2866\times 9.81$

$w=22.43\ N$

KengaRu [80]3 years ago

4 0

Answer:

Explanation:

Force, F = 12 N

distance, s = 16.4 m

time, t = 2.5 s

initial velocity, u = 0 m/s

Let a be the acceleration

use second equation of motion

s = ut + 1/2 at²

16.4 = 0 + 0.5 x a x 2.5 x 2.5

a = 5.25 m/s²

Let m be the mass of the tool.

F = ma

12 = m x 5.25

m = 2.286 kg

Now for vertical motion

h = 10.3 m

u = 0 m/s

t = 2.88 m/s²

Let the acceleration due to gravity on that planet is g'

Use second equation of motion

10.3 = 0 + 0.5 x g' x 2.88 x 2.88

g' = 2.485 m/s²

Let W' be the weight of the tool on that planet

W' = m g' = 2.286 x 2.485 = 5.68 N

Let W be the weight of the tool on the earth

W = m x g = 2.286 x 9.8 = 22.4 N

You might be interested in

A 10.0g piece of copper wire, sitting in the sun reaches a temperature of 80.0 C. how many Joules are released when the copper c

Zolol [24]

Answer:

150.8 J

Explanation:

The heat released by the copper wire is given by:

$Q=mC_s \Delta T$

where:

m = 10.0 g is the mass of the wire

Cs = 0.377 j/(g.C) is the specific heat capacity of copper

$\Delta T=40.0 C - 80.0 C=-40.0 C$ is the change in temperature of the wire

Substituting into the equation, we find

$Q=(10.0 g)(0.377 J/gC)(-40.0^{\circ})=-150.8 J$

And the sign is negative because the heat is released by the wire.

6 0

3 years ago

A hollow sphere of inner radius 8.82 cm and outer radius 9.91 cm floats half-submerged in a liquid of density 948.00 kg/m^3. (a)

kari74 [83]

Answer:

a) 0.568 kg

b) 474 kg/m³

Explanation:

Given:

Inner radius = 8.82 cm = 0.0882 m

Outer radius = 9.91 cm = 0.0991 m

Density of the liquid = 948.00 Kg/m³

a) The volume of the sphere = $\frac{4\pi}{3}\times(0.0991^2-0.0882^2)$

or

volume of sphere = 0.0012 m³

also, volume of half sphere = $\frac{\textup{Total volume}}{\textup{2}}$

or

volume of half sphere = $\frac{\textup{0.0012}}{\textup{2}}$

or

Volume of half sphere =0.0006 m³

Now, from the Archimedes principle

Mass of the sphere = Weight of the volume of object submerged

or

Mass of the sphere = 0.0006× 948.00 = 0.568 kg

b) Now, density = $\frac{\textup{Mass}}{\textup{Volume}}$

or

Density = $\frac{\textup{0.568}}{\textup{0.0012}}$

or

Density = 474 kg/m³

8 0

2 years ago

A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet

lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

<em>Current flowing in the wire, I = 4.00 mA</em>
<em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
<em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
<em>Length of wire, L = 2.00 m</em>
<em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>

<em />

The initial area of the copper wire;

$A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2$

The final area of the copper wire;

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2$

The initial drift velocity of the electrons is calculated as;

$v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s$

The final drift velocity of the electrons is calculated as;

$v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s$

The change in the mean drift velocity is calculated as;

$\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s$

The time of motion of electrons for the initial wire diameter is calculated as;

$t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s$

The time of motion of electrons for the final wire diameter is calculated as;

$t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s$

The average acceleration of the electrons is calculated as;

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2$

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0

2 years ago

Object 1 has a momentum of 10 kg m/s and Object 2 has a momentum of 25 kg m/s. Will it be easier to change the direction of move

svetoff [14.1K]

Answer:

I think its object 1

Explanation:

Because the object that has more weight has a greater momentum and the lightest object that has a less momentum will be easier to change because its lighter.

5 0

3 years ago

Friction occurs when the ____ and ____ of two surfaces grind against each other.

monitta

Friction occurs when the surfaces and heat of two surfaces grind against each other.

I think these are the answers

7 0

2 years ago