Ask question

Ask question

All categories

2 years ago

7

On her way home from work, Brenda drove 20 miles at 60 miles per hour. Due to poor weather conditions, she then reduced her spee

d to 30 miles per hour for the remainder of the trip. Her entire commute to work is 80 miles in distance. Based on the information provided above, which numerical value best represents how much less time would it have taken Brenda had she driven the entire distance at 40 miles per hour?
A. 1/12 hour
B. 1/6 hour
C. 1/4 hour
D. 1/3 hour
E. 2/3 hour

1 answer:

rjkz [21]2 years ago

3 0

Answer:

D

Explanation:

Speed = distance / time

her time for the first journey = 20 miles / 60 miles/hr = 1/3 hr

her time for second part of the journey = her remaining distance / her speed = (80 - 20) miles / 30 miles/hr = 60 miles / 30 miles/hr = 2 hrs

total time spend by her = 2 hr+ 1/3 hr = 2 1/3 hrs

her traveling the distance at 40 miles per hour = 80 miles / 40 miles /hr = 2 hrs

the time less she would drive if she drive the entire distance at 40 miles/hr = 2 1/3 hrs - 2 hrs = 1/3 hr

You might be interested in

Determine the magnitude and sense (direction) of the current in the 500-latex: \omega ω resistor when i = 30 ma.

VARVARA [1.3K]

Complete Question:

Check the circuit in the file attached to this solution

Answer:

Total current = 0.056 A(From left to right)

Explanation:

Let the current in loop 1 be I₁ and the current in loop 2 be I₂

Applying KVL to loop 1

30 - (I₁ - I₂)500 + I₂R + 15 = 0

45 - 500I₁ - 500I₂ + RI₂ = 0

I₁ = 30mA = 0.03 A

45 - 500(0.03) - 500I₂ + RI₂ = 0

30 -500I₂ + RI₂ = 0...............(1)

Applying kvl to loop 2

-RI₂ - 15 + 10 - 400I₁ = 0

-RI₂ = 5 + 400*0.03

RI₂ = -17 ................(2)

Put equation (2) into (1)

30 -500I₂ -17 = 0

-500I₂ = 13

I₂ = -13/500

I₂ = -0.026 A

The total current in the 500 ohms resistor = I₁ - I₂ = 0.03+0.026

Total current = 0.056 A

The current will flow from left to right

5 0

2 years ago

Ugonna stands at the top of an incline and pushes a 100−kg crate to get it started sliding down the incline. The crate slows to

Anna [14]

Answer:(a)891.64 N

(b)0.7

Explanation:

Mass of crate $m=100 kg$

Crate slows down in $s=1.5 m$

initial speed $u=1.77 m/s$

inclination $\theta =30^{\circ}$

From Work-Energy Principle

Work done by all the Forces is equal to change in Kinetic Energy

$W_{friction}+W_{gravity}=\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2$

$W_{gravity}=mg(0-h)=mgs\sin \theta$

$W_{gravity}=-mgs\sin \theta$

$W_{gravity}=-100\times 9.8\times 1.5\sin 30=-735 N$

change in kinetic energy $=\frac{1}{2}\times 100\times 1.77^2=156.64 J$

$W_{friction}=156.64+735=891.645$

(b)Coefficient of sliding friction

$f_r\cdot s=W_{friciton}$

$891.645=f_r\times 1.5$

$f_r=594.43 N$

and $f_r=\mu mg\cos \theta$

$\mu 100\times 9.8\times \cos 30=594.43$

$\mu =0.7$

5 0

2 years ago

A mover pushes a 255 kg piano

faust18 [17]

Answer:

$0.495 ms^{-2}$

Explanation:

According to the newton's second law of motion we can apply F=ma hear

Force = mass * acceleration

(assume the piano is moving left side )

←F = ma

$F_(pull)+ F_(push)= M*a\\77.5 + 48.7 = 255 *a\\a = 0.495 ms^{-2}$

7 0

2 years ago

Nisha was making a list of things that are obtained from plants and Amol was making a list of things that can dissolve in water.

Allisa [31]

Answer:

Sugar

Explanation:

Nisha's list: Amol's list

Jute Vinegar

Wood Lemon juice

Rubber Cooking soda

Nisha's list is made up of things that can be derived from plants. From the list given, oil, sugar and cotton can also be obtained from plants.

Most plant materials are organic matter.

Amol's list is made up of things that can dissolve in water. Sugar and table salt can also be dissolved in water.

Water is able to dissolve these materials because they are polar compounds. One rule of solubility is that like dissolves like.

New list:

Nisha's list: Amol's list

Jute Vinegar

Wood Lemon juice

Rubber Cooking soda

Sugar Sugar

Only sugar from the list can be added to both lists. It can be obtained from plant and can also dissolve in water.

6 0

2 years ago

Consider a double Atwood machine constructed as follows: A mass 4m is suspended from a string that passes over a massless pulley

kenny6666 [7]

Answer:

Hello your question is incomplete attached below is the complete question

Answer : x ( acceleration of mass 4m ) = $\frac{g}{7}$

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

Explanation:

Given data:

mass suspended = 4 meters

mass suspended at other end = 3 meters

first we have to express the kinetic and potential energy equations

The general kinetic energy of the system can be written as

T = $\frac{4m}{2} x^2 + \frac{3m}{2} (-x+y)^2 + \frac{m}{2} (-x-y)^2$

T = $4mx^2 + 2my^2 -2mxy$

also the general potential energy can be expressed as

U = $-4mgx-3mg(-x+y)-mg(-x-y)+constant=-2mgy +constant$

The Lagrangian of the problem can now be setup as

$L =4mx^2 +2my^2 -2mxy +2mgy + constant$

next we will take the Euler-Lagrange equation for the generalized equations :

Euler-Lagrange equation = $4x-y =0\\-2y+x +g = 0$

solving the equations simultaneously

x ( acceleration of mass 4m ) = $\frac{g}{7}$

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

8 0

2 years ago