All categories

3 years ago

Use the equation d = m/v [stack fraction), where d= density, m= mass, and v

Physics

1 answer:

sladkih [1.3K]3 years ago

5 0

Answer:

54.

Explanation:

6 (v) x 9 (m) = 54.

54 (m) / 6 (v) = 9 (d)

You might be interested in

What current is used to power the United States power grid?

alexdok [17]

The answer is Alternating Current

3 0

2 years ago

Read 2 more answers

A pitcher is in 85° of abduction, holding a 1.4 N baseball at point C, 65 cm from the joint axis at point O • The center of grav

erastovalidia [21]

I attached a Diagram for this problem.

We star considering the system is in equlibrium, so

Fm makes $90-(\theta+5)$ with vertical

Fm makes 70 with vertical

Applying summatory in X we have,

$\sum F_x = 0$

$W+1.4-Fm cos(70)$

We know that W is equal to

$W= 0.06*100N = 6N$

Substituting,

$Fm cos (70) = W+1.4N$

$Fm cos (70) = 6N + 1.4N$

$Fm = \frac{7.4}{cos(70)}$

$Fm = 21.636N$

<em>For the second part we know that the reaction force Fj on deltoid Muscle is equal to Fm, We can assume also that</em> $\beta = \theta$

3 0

2 years ago

A vehicle that goes from 5m/s to 45m/s in 8s. what is its acceleration?

GaryK [48]

Answer: 5m/s^2

Explanation:

V= 45m/s

U = 5m/s

t = 8s

a =?

V = u + at

45 = 5 + 8a

8a = 45 — 5

8a = 40

a = 40 / 8

a = 5m/s^2

3 0

3 years ago

a projectile is shot horizontally from the edge of a cliff, 230m above the ground. the projectile lands 300m from base of the cl

bekas [8.4K]

Answer:

The time taken by the projectile to hit the ground is 6.85 sec.

Explanation:

Given that,

Vertical height of cliff = 230 m

Distance = 300 m

Suppose, determine the time taken by the projectile to hit the ground.

We need to calculate the time

Using second equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Where, s = vertical height of cliff

u = initial vertical velocity

g = acceleration due to gravity

Put the value in the equation

$230=0+\dfrac{1}{2}\times9.8\times t^2$

$t=\sqrt{\dfrac{230\times2}{9.8}}$

$t=6.85 sec$

Hence, The time taken by the projectile to hit the ground is 6.85 sec.

7 0

3 years ago

Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p

Marta_Voda [28]

Answer:

Acceleration, $a=9.91\times 10^{15}\ m/s^2$

Explanation:

It is given that,

Separation between the protons, $r=3.73\ nm=3.73\times 10^{-9}\ m$

Charge on protons, $q=1.6\times 10^{-19}\ C$

Mass of protons, $m=1.67\times 10^{-27}\ kg$

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

$ma=\dfrac{kq^2}{r^2}$

$a=\dfrac{kq^2}{mr^2}$

$a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.67\times 10^{-27}\times (3.73\times 10^{-9})^2}$

$a=9.91\times 10^{15}\ m/s^2$

So, the acceleration of two isolated protons is $9.91\times 10^{15}\ m/s^2$ . Hence, this is the required solution.

3 0

3 years ago