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3 years ago

What is the mass of a large dog that weighs 441 newtons

1 answer:

5 0

Weight, w = mg. g ≈ 9.8 m/s². m = mass in kg. w is weight in N

441 N = m* 9.8

9.8m = 441

m = 441/9.8

m = 45 kg.

Mass of the dog is = 45 kg

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Two blocks of ice, one four times as heavy as the other, are at rest on a frozen lake. A person pushes each block the same dista

ANEK [815]

Answer:

The kinetic energy of the heavier block after the push is equal to the kinetic energy of the lighter block

Explanation:

Kinetic energy of smaller object

$K=(1/2)mv^2$

where m= mass of smaller object and v= velocity of smaller object

Also, it is given that heavier object is four times the mass of lighter object and consider its velocity as V

kinetic energy of heavier block $K '= (1/2) (4m) V^2$

Now, For smaller block , $v^2 - u^2=2aS$

[by Newtons laws of motion]

Also, $v^2 = 2(F/m)S$

Where S= displacement, F= force, u= initial velocity

So, $K=(1/2)m[2(F/m)]S$

$\Rightarrow K = FS$

For heavier block ,

$V^2 - u^2 = 2a'S$

or, $V^2=2(F /4m)S$

So, $K '= (1/2)(4m)[2(F/4m)S$

$\Rightarrow K'= FS$

Therefore the kinetic energy of the heavier block after the push is equal to the kinetic energy of the lighter block

5 0

3 years ago

If the wavelength is 10 m and the frequency is 5 Hz, what is the wave speed?

timama [110]

V = f x wavelength
V = 5 x 10
V = 50m/s

4 0

3 years ago

Momentum usually has the symbol p It is a vector, What is the correct way to

Nadya [2.5K]

Answer:

Add an arrow above the symbol p to show it is a vector. Sometimes it is italicized in textbooks.

Explanation:

6 0

3 years ago

Three-point charges are arranged on a line. Charge q3 = +5.00nC and is at the origin. Charge q2 = -3.00 nC and is at x = 5.00 cm

Anon25 [30]

Answer:

Explanation:

Given

Charge $q_3=+5\ nC$ is placed at origin

Charge $q_2=-3\ nC$ is placed at $x=5\ cm$

Charge $q_1$ is Placed at $x=2.5\ cm$

charge $q_1$ must be positive so as to balance the force on charge $q_3$

Force on $q_3$ due to $q_1$ is

$F_{31}=\frac{kq_1q_3}{r^2}$

here $r=2.5\ cm$

$F_{31}=\frac{kq_1q_3}{(2.5)^2}$

Force on $q_3$ due to $q_2$ is

$F_{32}=\frac{kq_3q_2}{r^2}$

here $r=5\ cm$

$F_{32}=\frac{kq_3q_2}{(5)^2}$

$F_{31}=F_{32}$

$\frac{kq_1q_3}{(2.5)^2}=\frac{kq_3q_2}{(5)^2}$

$q_1=q_2(\frac{2.5}{5})^2$

$q_1=3\times (\frac{2.5}{5})^2$

$q_1=3\times \frac{1}{4}$

$q_1=0.75\ nC$

5 0

4 years ago

Consider the program below:public class Test{ public static void main(String[] args) {Int[] a;a = new int[10];for(int i = 0; i &

ivolga24 [154]

Answer:

c. 65

Explanation:

The output is 65.

An array of length 10 is created first. Then, the first for-loop fill the array with different values; The array element now become: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. The array element are generated using the equation a[i] = i + 2; so when i is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. i must be less than the array length (10).

a[0] = 0 + 2 = 2

a[1] = 1 + 2 = 3

a[2] = 2 + 2 = 4

a[3] = 3 + 2 = 5

a[4] = 4 + 2 = 6

a[5] = 5 + 2 = 7

a[6] = 6 + 2 = 8

a[7] = 7 + 2 = 9

a[8] = 8 + 2 = 10

a[9] = 9 + 2 = 11

result variable is declared and initialized to 0.

The second for-loop goes through the array and add individual element to result.

6 0

3 years ago