Which line represents an object that is in motion where the acceleration is zero and the velocity is to the north?

How would a continents climate change if it drifted closer to the equator?

otez555 [7]

The equator is the closest part of out planet that is closest to the sun, which means that the continents closest to the equator will in turn be warmer.

8 0

3 years ago

"A proton is placed in a uniform electric field of 2750 N/C. You may want to review (Page) . For related problem-solving tips an

tekilochka [14]

To solve this problem we will start from the definition of Force, as the product between the electric field and the proton charge. Once the force is found, it will be possible to apply Newton's second law, and find the proton acceleration, knowing its mass. Finally, through the linear motion kinematic equation we will find the speed of the proton.

PART A ) For the electrostatic force we have that is equal to

$F=qE$

Here

q= Charge

E = Electric Force

$F=(1.6*10^{-19}C)(2750N/C)$

$F = 4.4*10^{-16}N$

PART B) Rearrange the expression F=ma for the acceleration

$a = \frac{F}{m}$

Here,

a = Acceleration

F = Force

m = Mass

Replacing,

$a = \frac{4.4*10^{-16}N}{1.67*10^{-27}kg}$

$a = 2.635*10^{11}m/s^2$

PART C) Acceleration can be described as the speed change in an instant of time,

$a = \frac{v_f-v_i}{t}$

There is not $v_i$ then

$a = \frac{v_f}{t}$

Rearranging to find the velocity,

$v_f = at$

$v_f = (2.635*10^{11})(1.4*10^{-6})$

$v_f = 3.689*10^{5}m/s$

7 0

3 years ago

An automobile engine takes in 4000 j of heat and performs 1100 j of mechanical work in each cycle. (a) calculate the engine's ef

Semmy [17]

(a) The efficiency of an engine is defined as the ratio between the work done by the engine and the heat it takes in:
$\eta= \frac{W}{Q_{in}}$
The engine in this problem does a work of $W=1100 J$ and it takes in $Q_{in}=4000 J$ of heat, therefore its efficiency is
$\eta= \frac{1100 J}{4000 J}=0.275 = 27.5 \%$

(b) The heat taken by the machine is 4000 J; of this amount of heat, only 1100 J are converted into useful work. This means that the rest of the heat is wasted. Therefore, the wasted heat is the difference between the heat in input and the work done by the engine:
$Q_{wasted}=Q_{in}-W=4000 J-1100 J=2900 J$

7 0

3 years ago

a 16.0 kg cart is being pulled by a 95.4 N force to the right and a 36.0 N force to the left. What is the acceleration of the ca

Inga [223]

Answer:

The cart's acceleration is $\approx 3.71\,\,\frac{m}{s^2}$