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Katen [24]
3 years ago
10

Whats the answer to 45 + 54 ÷ p - q

Physics
1 answer:
Lynna [10]3 years ago
4 0
There’s nothing to answer to
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I Need help with this problem i don’t know what to do
Mazyrski [523]

Answer:

The density of the sample is 36 g/cm³

Explanation:

m= 972g

l=3cm

V = l³ = 3³ = 27 cm³

density = mass/volume

= 972/27

= 36 g/cm³

8 0
3 years ago
Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400
hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

v^2=u^2+2as

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

v=u+at

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m

The beetle can jump to a height of <u>2.1 m</u>



7 0
3 years ago
What is the wavelength of a 6.00*10^2 Hz sound wave in air at 20C
Alex Ar [27]

Answer:

Solution

λ=v/n

Here, v=344 m s−1

n=22 MHz =22×106 Hz

λ=344/22×106=15.64×10−6m=15.64μm.

5 0
2 years ago
Benny has 20 jellybeans and wants to share with his friends how many will each friend get? there are 5 friends.
trasher [3.6K]

Answer each friend will get 3.33333 repeating if he is included. if only his friends are getting them then each one gets 4

Explanation:

devide 20/6 and 20/5 respectively.

7 0
3 years ago
An electron is released a short distance above earth's surface. a second electron directly below it exerts an electrostatic forc
slamgirl [31]

The mass of an electron is 9.109 x 10⁻³¹ kg

The weight of the electron is (mass) x (g) =  8.926 x 10⁻³⁰ Newton

The charge on an electron is -1.602 x 10⁻¹⁹ Coulomb

The repelling force between the two electrons is (K · q₁ · q₂ / r²) =

(8.98755 x 10⁹ N-m²/C²) x (1.602 x 10⁻¹⁹ C)² / D²

In order for the bottom one to just exactly hold the top one up at a distance 'D', the repelling force has to be exactly equal to the weight of the upper electron.

8.926 x 10⁻³⁰ N = (8.98755 x 10⁹ N-m²/C²)·(1.602 x 10⁻¹⁹ C)² / D²

We have to solve THAT ugly mess for ' D '.

Clean up the units first:

Cancel the C² on the right side, then divide each side by Newton:

8.926 x 10⁻³⁰ = (8.98755 x 10⁹ m²) x (1.602 x 10⁻¹⁹)² / D²

Now, let's multiply both sides by (D² x 10²⁹) :

D² x 8.926 x 10⁻¹ = (8.98755 m²) x (1.602)²

Divide each side by (0.8926):

D² = (8.98755 x 1.602²) / (0.8926)  meter²

D² = 25.84 m²

Take the square root of each side:

<em>D = 5.08 meters</em>

I am shocked, impressed, and amazed !

Are you shocked, impressed, or amazed ?

3 0
3 years ago
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