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Yuri [45]
3 years ago
6

Which question should Trudy write

Physics
1 answer:
Tanzania [10]3 years ago
6 0

A real question, one with a question mark?

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What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and
gayaneshka [121]

Force applied on the car due to engine is given as

F_1 = 300 N towards right

Also there is a force on the car towards left due to air drag

F_2 = 150 N towards left

now the net force on the car will be given as

\vec F_{net} = \vec F_1 + \vec F_2

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

F_{net} = F_1 - F_2

F_{net} = 300 - 150

F_[net} = 150 N

So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.

5 0
3 years ago
What is less dense water or oil steel or water helium or air or oil or water
Leni [432]

Answer:

air because their is nothing contained within the air other than all the solutions that you have listed

Explanation:

4 0
3 years ago
Read 2 more answers
Lord Beckett and members of the EIT Co. spot the Black Pearl in the distance making its way towards land. As a
KATRIN_1 [288]

Answer:

Explanation:

The cannonball goes a horizontal  distance of 275 m . It travels a vertical distance of 100 m

Time taken to cover vertical distance =  t ,

Initial velocity u = 0

distance s = 100 m

acceleration a = 9.8 m /s²

s = ut + 1/2 g t²

100 = .5 x 9.8 x t²

t = 4.51 s

During this time it travels horizontally also uniformly so

horizontal velocity Vx = horizontal displacement / time

= 275 / 4.51 = 60.97 m /s

Vertical velocity Vy

Vy = u + gt

= 0 + 9.8 x 4.51

= 44.2 m /s

Resultant velocity

V = √ ( 44.2² + 60.97² )

= √ ( 1953.64 + 3717.34 )

= 75.3 m /s

Angle with horizontal Ф

TanФ = Vy / Vx

= 44.2 / 60.97

= .725

Ф = 36⁰ .

6 0
3 years ago
ANSWER QUICK 30 POINTS
Lelu [443]
What do you need help on
4 0
3 years ago
A motorcyclist heading east through a small town accelerate at constant 4.0meter per seconds square after he leaves the limits.
SVETLANKA909090 [29]

A) The position at t = 2.0 sec is 43.0 m east

B) The position is 55 m east

Explanation:

A)

In order to solve the problem, we take the east direction as positive direction.

We know that:

- at t = 0, the motorcyclist is at a position of x_0 = 5.0 m

- at t = 0, the initial velocity of the motorcyclist is v_0 = 15.0 m east

- The acceleration of the motorcyclist is constant and it is a=4.0 m/s^2

Since the motion is a uniformly accelerated motion, the position of the motorcylist is given by the expression

x(t)=x_0 + v_0t + \frac{1}{2}at^2

where t is the time.

Substituting t = 2.0 s, we find the position:

x(2.0)=(5.0)+(15)(2.0)+\frac{1}{2}(4.0)(2.0)^2=43 m

B)

The velocity of the motoryclist can be found by calculating the derivative of the position. Therefore, it is:

v(t)=x'(t)=v_0 + at

where:

v_0=15.0 m/s is the initial velocity

a=4.0 m/s^2 is the acceleration

We want to find the time t at which the velocity is

v = 25 m/s

Solving the equation for t,

t=\frac{v-v_0}{a}=\frac{25-15}{4}=2.5 s

And therefore, the position at t = 2.5 s is:

x(2.5s)=5.0+(15.0)(2.5)+\frac{1}{2}(4)(2.5)^2=55 m

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

3 0
3 years ago
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