Ask question

Ask question

All categories

1 year ago

10

On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he

finds that, under the same circumstances, the stone returns to his hand in 2T. In both cases, he throws the stone with the same initial velocity and it feels negligible air resistance. The acceleration due to gravity on planet X (in terms of g) is _______.a) g/. b) g/4. c) 2g. d) g/2. e) g.

1 answer:

Liula [17]1 year ago

4 0

Answer:

correct is d) a ’= g / 2

Explanation:

For this exercise let's use the kinematics equations

On earth

v = v₀ - a t

a = (v₀- v) / T

On planet X

v = v₀ - a' t’

a ’= (v₀-v) / 2T

Let's substitute the land values in plot X

a’= a / 2

Now let's use Newton's second law

W = ma

m g = m a

a = g

We substitute

a ’= g / 2

So we see that on planet X the acceleration is half the acceleration of Earth's gravity

You might be interested in

What are the magnitude and direction of the force the pitcher exerts on the ball? (enter your magnitude to at least one decimal

murzikaleks [220]

Details are missing in the question. Complete text of the problem:

"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.

(a) Through what distance does it move before its release? (m)
(b) What are the magnitude and direction of the force the pitcher exerts on the ball? (Enter your magnitude to at least one decimal place.)"

Solution

(a) The pitcher accelerates the baseball from rest to a final velocity of $v_f = 16.5 m/s$ , so $\Delta v=16.5 m/s$ , in a time interval of $\Delta t = 181 ms=0.181 s$ . The acceleration of the ball in the horizontal direction (x-axis) is therefore

$a_x = \frac{\Delta v}{\Delta t}= \frac{16.5 m/s}{0.181 s}=91.2 m/s^2$

And the distance covered by the ball during this time interval, before it is released, is:

$S= \frac{1}{2} a_x (\Delta t)^2 = \frac{1}{2} (91.2 m/s^2)(0.181 s)^2=1.49 m$

(b) For this part we need to consider also the weight of the ball, which is $W=mg=2.28 N$

From this, we find its mass: $m= \frac{W}{g}= \frac{2.28 N}{9.81 m/s^2}=0.23 Kg$

Now we can calculate the magnitude of the force the pitcher exerts on the ball. On the x-axis, we have

$F_x = m a_x = (0.23 kg)(91.2 m/s^2)=20.98 N$

We also know that the ball is moving straight horizontally. This means that the vertical component of the force exerted by the pitcher must counterbalance the weight of the ball (acting downward), in order to have a net force of zero along the y-axis, and so:

$F_y=W=mg=2.28 N$ (upward)

So, the magnitude of the force is

$F= \sqrt{F_x^2+F_y^2}= \sqrt{(20.98N)^2+(2.28N)^2}=21.2 N$

To find the direction, we should find the angle of F with respect to the horizontal. This is given by

$\tan \alpha = \frac{F_y}{F_x}= \frac{2.28 N}{20.98 N}=0.11$

From which we find $\alpha=6.2^{\circ}$

7 0

2 years ago

Read 2 more answers

What is an example of convection currents? a. marshmallows toasting over a campfire b. a pot being heated by an electric burner,

SCORPION-xisa [38]

Answer:

B

Explanation:

because, convection is the transfer of heat between fluid substances/materials

7 0

2 years ago

Read 2 more answers

Imagine you derive the following expression by analyzing the physics of a particular system: v2=v20+2ax. The problem requires so

lisabon 2012 [21]

As per kinematics equation we are given that

$v^2 = v_o^2 + 2ax$

now we are given that

a = 2.55 m/s^2

$v_0 = 21.8 m/s$

$v = 0$

now we need to find x

from above equation we have

$0^2 = 21.8^2 + 2(2.55)x$

$0 = 475.24 + 5.1 x$

$x = 93.2 m$

so it will cover a distance of 93.2 m

7 0

2 years ago

Read 2 more answers

Nastasia [14]

Answer:

8 cm

Ladder is safer than to walk on it

Explanation:

Streched = Final length - initial length

= 66 - 58

= 8 cm

When your are walking along frozen ice pond your weight is not distributed .It's acting on directly on the ice pond because of contact Area is too small between pond and you. There for According to pressure equation ,

P = F/A

When area is less pressure is less.

If use ladder instead of going by foot. Contact area will be high between pond and ladder. There for pressure is reduced .

3 0

2 years ago

An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its

FromTheMoon [43]

Answer:

The terminal speed of this object is 12.6 m/s

Explanation:

It is given that,

Mass of the object, m = 80 kg

The magnitude of drag force is,

$F_{drag}=12v+4v^2$

The terminal speed of an object is attained when the gravitational force is balanced by the gravitational force.

$F_{drag}=mg$

$12v+4v^2=80\times 9.8$

$4v^2+12v=784$

On solving the above quadratic equation, we get two values of v as :

v = 12.58 m/s

v = -15.58 m/s (not possible)

So, the terminal speed of this object is 12.6 m/s. Hence, this is the required solution.

6 0

1 year ago