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3 years ago

If a bus travels 120 miles and travels at 35 miles per hour, how long will the trip take?

Physics

1 answer:

Artyom0805 [142]3 years ago

7 0

Speed = Distance/Time

we have speed and distance from the question;

Now,

35 = 120/t

35/120 = 1/t

120/35 = t

3.42 hours = t

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A fuel cell that is becoming very popular for cars runs on:

jok3333 [9.3K]

It runs on Hydrogen gas.

Actually, using hydrogen as a fuel is not new. We used to use it on air vehicle like air balloon. But back then, we still cannot figure out how to safely use this because Hydrogen exlodes rather easily

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3 years ago

In a certain region of space, the electric potential is V(x,y,z)=Axy-Bx^2+Cy , where A,B , and C are positive constants.a) Calcu

laiz [17]

Answer:

a) Eₓ = - A y + 2B x , b) Ey = -Ax –C , c) Ez = 0 , d) The correct answer is 3

Explanation:

The electric field and the electric power are related

E = - dV / ds

a) Let's find the electric field on the x axis

Eₓ = - dV / dx

dV / dx = A y - B 2x

Eₓ = - A y + 2B x

b) calculate the electric field on the y-axis

Ey = - dV / dy

dV / dy = A x + C

Ey = -Ax –C

c) the electric field on the z axis

dv / dz = 0

Ez = 0

.d) at which point the electric field is zero

Since the electric field is a vector quantity all components must be zero

X axis

0 = = - A y + 2B x

y = 2B / A x

Axis y

0 = -Ax –C

.x = -C / A

We substitute this value in the previous equation

.y = 2B / A (-C / A)

.y = 2 B C / A2

The correct answer is 3

6 0

3 years ago

A certain moving electron has a kinetic energy of 0.991 × 10−19 J. Calculate the speed necessary for the electron to have this e

alisha [4.7K]

Answer: The speed necessary for the electron to have this energy is 466462 m/s

Explanation:

Kinetic energy is the energy posessed by an object by virtue of its motion.

$K.E=\frac{1mv^2}{2}$

K.E= kinetic energy = $0.991\times 10^{-19}J$

m= mass of an electron = $9.109\times 10^{-31}kg$

v= velocity of object = ?

Putting in the values in the equation:

$0.991\times 10^{-19}J=\frac{1\times 9.109\times 10^{-31}kg\times v^2}{2}$

$v=466462m/s$

The speed necessary for the electron to have this energy is 466462 m/s

4 0

2 years ago

Read 2 more answers

An object begins x=75.2 m and undergoes a displacement of -48.7 m. what is its final position?

xz_007 [3.2K]

Answer:

26.5 m

Explanation:

$x_{o}$ = initial position of the object = 75.2 m

$x$ = final position of the object

$d$ = displacement of the object = - 48.7

Displacement of the object is given as the difference of final and initial position of the object

$d = x - x_{o}$

Inserting the values

- 48.7 = x - 75.2

x = 26.5 m

8 0

2 years ago

A group of students collected the data shown below while attempting to measure the coefficient of static friction (of course, it

anzhelika [568]

Answer:

0.130

Explanation:

From the given data, the coefficient of static friction for each trial are:

1. 0.053

2. 0.081

3. 0.118

4. 0.149

5. 0.180

6. 0.198

The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

= 0.779

So that;

the average coefficient of static friction = $\frac{sum of coefficient of static friction}{number of trials}$

= $\frac{0.779}{6}$

= 0.12983

The average coefficient of static friction is 0.130

8 0

2 years ago