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lys-0071 [83]
3 years ago
5

If a person is throwing darts at a target in shoots three in a row in the same spot on the 20 section outer ring, but none in th

e bull’s-eye, the person throwing could be described as _____.
1) accurate, but not exact
2) precise, not accurate
3) accurate, but not precise
4) exact, but not precise
Physics
2 answers:
Harman [31]3 years ago
8 0
2)precise,not accurate
Oksi-84 [34.3K]3 years ago
5 0
The answer would be 2) precise, not accurate
You might be interested in
a 300kg motorboat is turned off as it approaches a dock and coasts towards it at .5 m/s. Isaac, whose mass is 62 kg jumps off th
Zolol [24]

-- Before he jumps, the mass of (Isaac + boat) = (300 + 62) = 362 kg,
their speed toward the dock is 0.5 m/s, and their linear momentum is

  Momentum = (mass) x (speed) = (362kg x 0.5m/s) = <u>181 kg-m/s</u>

<u>relative to the dock</u>. So this is the frame in which we'll need to conserve
momentum after his dramatic leap.

After the jump:

-- Just as Isaac is coiling his muscles and psyching himself up for the jump,
he's still moving at 0.5 m/s toward the dock.  A split second later, he has left
the boat, and is flying through the air at a speed of 3 m/s relative to the boat.
That's 3.5 m/s relative to the dock.

    His momentum relative to the dock is (62 x 3.5) = 217 kg-m/s toward it.

But there was only 181 kg-m/s total momentum before the jump, and Isaac
took away 217 of it in the direction of the dock.  The boat must now provide
(217 - 181) = 36 kg-m/s of momentum in the opposite direction, in order to
keep the total momentum constant.

Without Isaac, the boat's mass is 300 kg, so 

                     (300 x speed) = 36 kg-m/s .

Divide each side by 300:  speed = 36/300 = <em>0.12 m/s ,</em> <u>away</u> from the dock.
=======================================

Another way to do it . . . maybe easier . . . in the frame of the boat.

In the frame of the boat, before the jump, Isaac is not moving, so
nobody and nothing has any momentum.  The total momentum of
the boat-centered frame is zero, which needs to be conserved.

Isaac jumps out at 3 m/s, giving himself (62 x 3) = 186 kg-m/s of
momentum in the direction <u>toward</u> the dock.

Since 186 kg-m/s in that direction suddenly appeared out of nowhere,
there must be 186 kg-m/s in the other direction too, in order to keep
the total momentum zero.

In the frame of measurements from the boat, the boat itself must start
moving in the direction opposite Isaac's jump, at just the right speed 
so that its momentum in that direction is 186 kg-m/s.
The mass of the boat is 300 kg so
                                                         (300 x speed) = 186

Divide each side by 300:  speed = 186/300 = <em>0.62 m/s</em>    <u>away</u> from the jump.

Is this the same answer as I got when I was in the frame of the dock ?
I'm glad you asked. It sure doesn't look like it.

The boat is moving 0.62 m/s away from the jump-off point, and away from
the dock.
To somebody standing on the dock, the whole boat, with its intrepid passenger
and its frame of reference, were initially moving toward the dock at 0.5 m/s.
Start moving backwards away from <u>that</u> at 0.62 m/s, and the person standing
on the dock sees you start to move away <u>from him</u> at 0.12 m/s, and <em><u>that's</u></em> the
same answer that I got earlier, in the frame of reference tied to the dock.

  yay !

By the way ... thanks for the 6 points.  The warm cloudy water
and crusty green bread are delicious.


4 0
3 years ago
Which kind of force do you exert on an object when you pull it toward you?
jarptica [38.1K]

By definition we have to:

Applied force: It is the external force that acts directly on a body.

Therefore, we can say that if you have an object and push it towards yourself, you are exerting an external force on the object.

This external force was not acting on the object previously, therefore, it is a force that you are applying at that moment.

Answer:

you exert an Applied Force on an object when you pull it towards you

A. Applied Force

4 0
3 years ago
Read 2 more answers
If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a
PolarNik [594]

This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

Angular Velocity at time 0s = 12 rad/s

Angular Velocity at time 0.15s = 24 rad/s

a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

Answer:

a) the angular acceleration is 80 rad/s²

b) the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s w_o = 12 rad/s

Angular Velocity at time 0.15s w_f = 24 rad/s

a) What is the angular acceleration;

Angular acceleration ∝ = ( w_f - w_o ) / dt

we substitute

Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

b)

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

8 0
2 years ago
It is correct to say that impulse is equal toA) momentum.B) the change in momentum.C) the force multiplied by the distance the f
Elena-2011 [213]

Answer:

B) the change in momentum.

Explanation:

The impulse is defined as the product between the force applied on an object (F) and the duration of the collision (\Delta t):

J=F \Delta t (1)

We can rewrite the force by using Newton's second law, as the product between mass (m) and acceleration (a):

F=ma

So, (1) becomes

J=ma \Delta t

Now we can also rewrite the acceleration as ratio between the change in velocity and change in time: a=\frac{\Delta v}{\Delta t}. If we substitute into the previous equation, we find

J=m\frac{\Delta v}{\Delta t}\Delta t=m\Delta v

And the quantity m\Delta v is equivalent to the change in momentum, \Delta p.

6 0
3 years ago
A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes
Nitella [24]

Answer:

The value of F= - 830 N

Since the force is negative, it implies direction of the force applied was due south.

Explanation:

Given data:

Mass = 1000-kg

Distance, d = 240 m

Initial velocity, v1 = 20.0 m/s

Final velocity, v2 = 0 (since the car came to rest after brake was applied)

v2²= v1² + 2ad (using one of the equation of motion)

0=  20² + (2 x a x  240)

0= 400 + 480 a

a = - 400/480

a = - 0.83 m/s²

Then, imputing the value of a into

F = ma

F = 1000 kg x ( - 0.83 m/s²)

F= - 830 N

The car was driving toward the north, and since the force is negative, it implies direction of the force applied was due south.

3 0
2 years ago
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