Ask question

Ask question

All categories

3 years ago

12

f an arrow is shot upward on the moon with velocity of 35 m/s, its height (in meters) after t seconds is given by h(t)=35t−0.83t

2. (a) Find the velocity of the arrow after 3 seconds.

1 answer:

inysia [295]3 years ago

7 0

Answer:

The velocity of the arrow after 3 seconds is 30.02 m/s.

Explanation:

It is given that,

An arrow is shot upward on the moon with velocity of 35 m/s, its height after t seconds is given by the equation:

$h(t)=35t-0.83t^2$

We know that the rate of change of displacement is equal to the velocity of an object.

$v(t)=\dfrac{dh(t)}{dt}\\\\v(t)=\dfrac{d(35t-0.83t^2)}{dt}\\\\v(t)=35-1.66t$

Velocity of the arrow after 3 seconds will be :

$v(t)=35-1.66t\\\\v(t)=35-1.66(3)\\\\v(t)=30.02\ m/s$

So, the velocity of the arrow after 3 seconds is 30.02 m/s. Hence, this is the required solution.

You might be interested in

Kate is working on a project in her tech education class. She plans to assemble a fan motor. Which form of energy does the motor

andreyandreev [35.5K]

Motion Energy

I am writing this so it can be more than 20 letters

7 0

4 years ago

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

A tennis ball moves 16 meters northward, then 22 meters southward, then 12 meters northward, and finally 32 meters southward.

igomit [66]

U didn’t ask a question so I’ll just say that’s cool that it moves like that

8 0

3 years ago

A furniture shop refinishes chairs. Employees use one of two methods to refinish each chair. Method I takes 0.5 hours and the ma

satela [25.4K]

Answer:

method I = 88 chairs

method II = 62 chairs

Explanation:

This problem can be modeled by a system of two linear equations.

Define x as the number of chairs refinished by method I and y by method II

The sum of hours spent on both methods should equal 199 and the sum of total material cost should equal $1226, therefore:

$0.5x + 2.5y = 199$

$9x+7y = 1226$

Multiplying the first equation by -18 and adding it to the second equation we can solve for the value of y:

$9x+7y +(-9x - 45)= 1226+(-3582)\\y=\frac{2356}{38} = 62\\$

We can now apply the value of y found to the first equation and solve for x:

$0.5x+2.5*62 = 199\\x=\frac{199 - (2.5*62)}{0.5} = 88$

Therefore, they should refinish 88 chairs with method I and 62 chairs with method II

4 0

3 years ago

Acceleration due to gravity is

Korolek [52]

Acceleration due to gravity is different in every location, because gravity itself is different in every location.

Here are a few values of gravitational acceleration in various places:

-- Surface of Jupiter . . . 24.8 m/s²

-- Surface of Mars . . . 3.7 m/s²

-- Surface of the Sun . . . 274 m/s²

-- Surface of Earth . . . 9.8 m/s²

-- In orbit 300 miles above the Earth's surface . . . 8.5 m/s²

-- Surface of Earth's Moon . . . 1.6 m/s²

3 0

4 years ago