A force is applied to a 13.0 kg wagon to pull it down the sidewalk at a constant speed. The coefficient of

The brakes on a 15,680 N car exert a stopping force of 640 N.

Murrr4er [49]

Answer:

1568 Kg is the answer

Explanation:

6 0

3 years ago

A wire of cross-sectional area 5.00 106 m2 has a resistance of 1.75 O. What is the resistance of a wire of the same material and

MakcuM [25]

Answer:

the resistance of the second wire is 1 ohm.

Explanation:

Given;

cross sectional area of the first wire, A₁ = 5.00 x 10⁶ m²

resistance of the first wire, R₁ = 1.75 ohms

cross sectional area of the second wire, A₂ = 8.75 x 10⁶ m²

resistance of the second wire, R₂ = ?

The resistance of a wire is given as;

R ∝ $\frac{L}{A}$

Since the length of the two wires is constant

R₁A₁ = R₂A₂

$R_2 = \frac{R_1A_1}{A_2} \\\\R_2 = \frac{1.75\ \times \ 5.00\times 10^6}{8.75\times 10^6} \\\\R_2 = 1 \ ohm$

Therefore, the resistance of the second wire is 1 ohm.

6 0

3 years ago

300N effort is applied to lift the load of 900N by using a first class lever. If the distance between the load and fulcrum is 20

Klio2033 [76]

Answer:

Effort=300N

Load=900N

Load distance=20m

Now,

Ed=(L*Ld)/E

Ed=(900*20)/300

Ed=1800/300

Ed=6m

Explanation:

7 0

2 years ago

Read 2 more answers

What type of relationship exists between momentum and mass?

alina1380 [7]

Answer:

A. Direct

Explanation:

Thats just how it be

4 0

3 years ago

Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k

larisa86 [58]

Answer:

part (a) $a_1\ =\ 2.9\ kg$

Part (b) $a_2\ =\ 6.25\ kg$

Explanation:

Given,

Mass of the larger disk = $M_2\ =\ 1.60\ kg$
Mass of the smaller disk = $M_1\ =\ 0.900\ kg$
Radius of the larger disk = $R_2\ =\ 5.00\ cm\ =\ 0.05\ m$
Radius of the smaller disk = $R_1\ =\ 2.45\ cm\ =\ 0.0245\ m$
Mass of the block = M = 1.60 kg

Both the disks are welded together, therefore total moment of inertia of the both disks are the summation of the individual moment of inertia of the disks.

$\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}M_1R_1^2\ +\ \dfrac{1}{2}M_2R_2^2\\\Rightarrow I\ =\ \dfrac{1}{2} (0.9\times 0.0245^2\ +\ 1.60\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2$

part (a)

Given that a block of mass m which is hanging with the smaller disk,

Let 'T' be 'a' be the tension in the string and acceleration of the block.

From the free body diagram of the smaller block,

$mg\ -\ T\ =\ ma\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,eqn (1)$

From the pulley,

$\sum \tau\ =\ I\alpha\\\Rightarow T\times R_1\ =\ I\alpha\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{I\alpha}{R_1^2}\,\,\,eqn(2)$

From the equation (1) and (2),

$mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.0245^2}}\ +\ 1.60}\\\Rightarow a\ =\ 2.91\ m/s^2$

part (b)

Above expression for the acceleration of the block is only depended on the radius of the pulley.

Radius of the larger pulley = $R_2\ =\ 0.05\ m$

Let $a_2$ be the acceleration of the block while connecting to the larger pulley. $\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_2^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.05^2}\ +\ 1.60}}\\\Rightarow a\ =\ 6.25\ m/s^2$

4 0

3 years ago