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4 years ago

6

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

2 answers:

taurus [48]4 years ago

7 0

the answer to the potential energy of the baby and stroller is 1960 J

Alekssandra [29.7K]4 years ago

3 0

so, question number 10 answer is 82 watts

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A golfer tees off from the top of a rise, giving the golf ball an initial velocity of 44 m/s at an angle of 25° above the horizo

mars1129 [50]

Answer:

a) $y=17.6m$

b) $v_{x}=39.87m/s\\v_{y}=-28.05m/s$

Explanation:

From the exercise we got our <u>initial data</u>

$v_{o}=44m/s\\\beta =25\\x=190m$

a) To find <u><em>maximum height</em></u> we know that at that point $v_{y}=0$

$v_{y} ^{2} =v_{oy} ^{2} +2a(y-y_{o} )$

$0=(44sin25)^{2} -2(9.8)y$

Solving for y

$y=\frac{(44sin25)^{2} }{2(9.8)} =17.6 m/s$

b) Since we know that the ball strikes the fairway 190 m away

$x=v_{ox}t\\ 190=44cos25t$

Solving for t

$t=4.76s$

Now, we can calculate the speed of the ball in both axes

$v_{x}=44cos25=39.87m/s$

$v_{y}=v_{oy}+at$

$v_{y}=44sin25-(9.8)(4.76)=-28.05m/s$

The <em>negative</em> sign means the direction of the ball at that point.

5 0

4 years ago

A 49 kg person is being dragged in their sleeping bag to the lake by a 593 N

Alex777 [14]

Answer:

485.62 N

Explanation:

To obtain the magnitude of unbalanced forces acting on the body;

Unbalanced Force = Horizontal Component of Applied Force - Frictional Force

Frictional Force = Horizontal Component of Applied Force - Unbalanced Force

f = frictional force = ?

F = Applied force = 593 N

m = mass of person = 49 kg

a = acceleration = 0.57 m/s²

θ = Angle with horizontal = 30°

Hence;

Horizontal Component of Applied Force = (593 N)(Cos 30°)

Unbalanced Force = (49 kg)(0.57 m/s²)

f = (593 N)(Cos 30°) - (49 kg)(0.57 m/s²)

f = 513.55 N - 27.93 N

f = 485.62 N

6 0

3 years ago

A racecar driver is driving her car down the drag strip at 120 m/s. What is the shortest distance in which she can brake and sto

DaniilM [7]

Answer:

1034.78 m

Explanation:

The shortest distance is the displacement of the car from initial position to final position.

Displacement of body is given using Newton's equations of motion.

Given:

Initial velocity, $u = 120$ m/s

Final velocity, $v = 0$ m/s( As the car stops in its final position)

Coefficient of static friction, $\mu=0.71$

Acceleration due to gravity, $g=9.8\textrm{ } m/s^{2}$

Now, when brakes are applied, only friction acts on the body in a direction opposite to that of its motion.

The acceleration of the car when friction acts the stopping force is given as:

$a=- \mu g$ .

The acceleration is negative as it reduces the velocity of the motion and acts in the direction opposite to that of the motion.

Plug in 0.71 for $\mu$ and 9.8 m/s² for $g$ . Solve for $a$ .

So, $a=-0.71\times 9.8=-6.958\textrm{ }m/s^{2}$ .

Now, displacement of the car is given using the following equation of motion:

$v^{2}=u^{2} +2aS$

Here, $S$ is the displacement of the racing car.

Plug in 120 m/s for $u$ , 0 m/s for $v$ , -6.958 m/s² for $a$ . Solve for $S$ . This gives,

$0^{2} =(120)^{2}+2(-6.958)S\\13.916S=14400\\S=\frac{14400}{13.916}=1034.78\textrm{ m}$

Therefore, the shortest distance in which she can brake and stop is 1034.78 m.

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4 years ago

Read 2 more answers

Four solid balls, each with a different mass, are moving at the same speed. Which ball would require the most force to stop its

zhannawk [14.2K]

Obviouly 20kg.............

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4 years ago

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3 years ago