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3 years ago

7

The number of perpandicular components of a force is

1 answer:

Arturiano [62]3 years ago

3 0

Answer. 2 components

one directed upwards, and the other directed rightwards

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Enter the expression 2gΔym−−−−−√, where Δ is the uppercase Greek letter Delta.

Anon25 [30]

This doesnt make sense. 2g<span>Δym is simply an equation. </span>

5 0

3 years ago

Given two vectors A--> = 4.20 i^+ 7.20 j^ and B--> = 5.70 i^− 2.40 j^ , find the scalar product of the two vectors A-->

Arisa [49]

Answer:

$\vec{A}\times \vec{B}=-51.12\hat{k}$

$\theta=83.2^{\circ}$

Explanation:

We are given that

$\vec{A}=4.2\hat{i}+7.2\hat{j}$

$\vec{B}=5.70\hat{i}-2.40\hat{j}$

We have to find the scalar product and the angle between these two vectors

$\vec{A}\times \vec{B}=\begin{vmatrix}i&j&k\\4.2&7.2&0\\5.7&-2.4&0\end{vmatrix}$

$\vec{A}\times \vec{B}=\hat{k}(-10.08-41.04)=-51.12\hatk}$ $\hat{k}$

Angle between two vectors is given by

$sin\theta=\frac{\mid a\times b\mid}{\mid a\mid \mi b\mid}$

Where $\theta$ in degrees

$\mid{\vec{A}}\mid=\sqrt{(4.2)^2+(7.2)^2}=8.3$

Using formula $\mid a\mid=\sqrt{x^2+y^2}$

Where x= Coefficient of unit vector i

y=Coefficient of unit vector j

$\mid{\vec{B}}\mid=\sqrt{5.7)^2+(-2.4)^2}=6.2$

$\mid{\vec{A}\times \vec{B}}\mid=\sqrt{(-51.12)^2}=51.12$

Using the formula

$sin\theta=\frac{51.12}{8.3\times 6.2}=0.993$

$\theta=sin^{-1}(0.993)=83.2$ degrees

Hence, the angle between given two vectors= $83.2^{\circ}$

7 0

4 years ago

A 9Ω resistance is cut into three equal parts and connected in parallel. Find the equivalent resistance of the combination.

sammy [17]

Answer:

the ans i 1 ohm

Explanation:

if i cut a resistor of 9 ohm into 3 equal parts then each resistor will have a resistance of 3 ohm and if they are in parallel combination the net resistance will be R

so 1/R=1/R1+1/R2+1/R3

1/R=1/3+1/3+1/3

R=1

7 0

3 years ago

The equivalent resistance of two resistors connected in series is always greater than the equivalent resistance of the same two

seraphim [82]

Answer:

True

Explanation:

Because the resistors in series is the sum of the two resistors given as

R= R1+R2

While that of resistors in parallel is the sum of the reciprocal of the resistance given as

1/R = 1/ R1+ 1/R2

So that of series connection will be greater

4 0

3 years ago

A horizontal 745 N merry-go-round of radius

Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is 544J

Explanation:

Given :

Weight w = 745 N

Radius r = 1.45 m

Force = 56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62 = ?

Solution:

Step 1: Finding the Mass of merry-go-round

$m = \frac{ weight}{g}$

$m = \frac{745}{9.81 }$

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I = $0.5 \times m \times r^2$

Substituting the values

Moment of Inertia of solid cylinder I

=> $0.5 \times 76.02 \times (1.45)^2$

=> $0.5 \times 76.02\times 2.1025$

=> $79.91 kg.m^2$

Step 3: Finding the Torque applied T

Torque applied T = $F \times r$

Substituting the values

T = $56.3 \times 1.45$

T = 81.635 N.m

Step 4: Finding the Angular acceleration

Angular acceleration , $\alpha = \frac{Torque}{Inertia}$

Substituting the values,

$\alpha = \frac{81.635}{79.91}$

$\alpha = 1.021 rad/s^2$

Step 4: Finding the Final angular velocity

Final angular velocity , $\omega = \alpha \times t$

Substituting the values,

$\omega = 1.021 \times 3.62$

$\omega = 3.69 rad/s$

Now KE (100% rotational) after 3.62s is:

KE = $0.5 \times I \times \omega^2$

KE = $0.5 \times 79.91 \times 3.69^2$

KE = 544J

6 0

4 years ago