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3 years ago

What is -37c on the Fahrenheit scale

Physics

1 answer:

Arisa [49]3 years ago

8 0

-35 is the ans.wer.hope this hels

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A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

xz_007 [3.2K]

Answer:

a) $\Delta{t} = 5.39s$

b) the motorcycle travels 155 m

Explanation:

Let $t_2-t_1 = \Delta{t}$ , then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

$v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}$

where:

$v_{m2}$ is the speed of the motorcycle at time 2

$v_{c}$ is the velocity of the car (constant)

$v_{0}$ is the velocity of the car and the motorcycle at time 1

d is the distance between the car and the motorcycle at time 1

x is the distance traveled by the car between time 1 and time 2

Solving the system of equations:

$\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]$

$v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s$

For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:

$x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933$

3 0

3 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 proton is projected in the positive x direction into a region of a uniform electric field →E =(-6.00 × 10⁵) i^ N/C at t=0 . Th

anzhelika [568]

The acceleration of the proton that is projected in the positive x direction into a region of a uniform electric field is 5.76×〖10〗^13 m/s^2

The product of the field's strength and the charge's strength yields the magnitude of the electric force acting on a charge traveling in a magnetic field region.

The electric force magnitude acting on the charge is expressed in the equation below.

F=|→E|×|q|

F=|-6.00×〖10〗^5 N/C|×||+1.602×〖10〗^(-19) C|

F=9.612×〖10〗^(-14) N

Newton's second law of motion states that the magnitude of a force is equal to the product of a proton's mass and its acceleration. Where the magnitude of the acceleration of the proton is ;

a=F/m

Where F is the force and m is the mass;

Inserting the values into the equation,

a=(9.612×〖10〗^(-14) N)/(1.67×〖10〗^(-27) kg)

a=5.76×〖10〗^13 m/s^2

Therefore, the acceleration of proton is 5.76×〖10〗^13 m/s^2 #SPJ4

brainly.com/question/13263306

#SPJ4

7 0

1 year ago

Find the magnitude of the vector v given its initial and terminal points. Round your answer to four decimal places.

Sliva [168]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is $|v| = 6.93$

Explanation:

From the question we are told that

The initial point is $(x_1 , y_1 , z_1 ) = (-1 , 7 , 4 )$

The terminal point is $(x_2 , y_2 , z_2) = (-5 , 11, 8 )$

Generally the magnitude of the vector is mathematically represented as

$|v| = \sqrt{(x_2 -x_1 )^2 + (y_2 - y_1 )^2 + (z_2 -z_1)^2}$

=> $|v| = \sqrt{(-5 -(-1) )^2 + (11 - 7 )^2 + (8 -4)^2}$

=> $|v| = 6.93$

3 0

3 years ago

How does the government involve in a child or young person’s upbringing in your<br> country?

dolphi86 [110]

The government are involved in a child or young person’s upbringing in my country by providing the right amenities and ensure the justice system is fair.

<h3>What is Upbringing?</h3>

This is defined as the way an individual treated at a young age by older adults such as parents.

Providing amenities will ensure the upbringing of the child is easier and a fair justice system will help to control crime.

Read more about Upbringing here brainly.com/question/12397063

#SPJ1

8 0

2 years ago