What is the density of a substance that has a mass of 20 g and a volume of 10mL

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

3 years ago

What is the frequency of a pendulum of length 1.50 m at a location where 1 point the acceleration due to gravity is 9.79 m/s^2?

ivann1987 [24]

Answer:

Explanation:The simple pendulum calculator finds the period and frequency of a ... Acceleration of gravity (g) ... Pendulum length (L) ... First of all, a simple pendulum is defined to be a point mass or bob (taking ... For example, it can be equal to 2 m. ... Find the frequency as the reciprocal of the period: f = 1/T = 0.352 Hz

4 0

3 years ago

PLEASE HELPPPP!!!!

Tresset [83]

you will hear a higher pitch due to a higher frequency.

3 0

3 years ago

Read 2 more answers

A current is induced by moving a magnet in and out of a coil of wire. What will happen to the induced current if you move the

Virty [35]

the answer is c. The induced current will increase

5 0

2 years ago

Consider the video you just watched. Suppose we replace the original launcher with one that fires the ball upward at twice the s

sveta [45]

Answer:

b

Explanation:

Given:

- The ball is fired at a upward initial speed v_yi = 2*v

- The ball in first experiment was fired at upward initial speed v_yi = v

- The ball in first experiment was as at position behind cart = x_1

Find:

How far behind the cart will the ball land, compared to the distance in the original experiment?

Solution:

- Assuming the ball fired follows a projectile path. We will calculate the time it takes for the ball to reach maximum height y. Using first equation of motion:

v_yf = v_yi + a*t

Where, a = -9.81 m/s^2 acceleration due to gravity

v_y,f = 0 m/s max height for both cases:

For experiment 1 case:

0 = v - 9.81*t_1

t_1 = v / 9.81

For experiment 2 case:

0 = 2*v - 9.81*t_2

t_2 = 2*v / 9.81

The total time for the journey is twice that of t for both cases:

For experiment 1 case:

T_1 = 2*t_1

T_1 = 2*v / 9.81

For experiment 2 case:

T_2 = 2*t_2

T_2 = 4*v / 9.81

- Now use 2nd equation of motion in horizontal direction for both cases:

x = v_xi*T

For experiment 1 case:

x_1 = v_x1*T_1

x_1 = v_x1*2*v / 9.81

For experiment 2 case:

x_2 = v_x2*T_2

x_2 = v_x2*4*v / 9.81

- Now the x component of the velocity for each case depends on the horizontal speed of the cart just before launching the ball. Using conservation of momentum we see that both v_x2 = v_x1 after launch. Since the masses of both ball and cart remains the same.

- Hence; take ratio of two distances x_1 and x_2:

x_2 / x_2 = v_x2*4*v / 9.81 * 9.81 / v_x1*2*v

Simplify:

x_1 / x_2 = 2

- Hence, the amount of distance traveled behind the cart in experiment 2 would be twice that of that in experiment 1.

3 0

2 years ago