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Goshia [24]
2 years ago
13

In a novel from 1866 the author describes a spaceship that is blasted out of a cannon with a speed of about 11.000 m/s. The spac

eship is approximately 274 m long but part of it is packed with gunpowder, so it accelerates over a distance of only 213 m. What was the acceleration (in m/s2) experienced by the occupants of the spaceship during the launch?
Physics
1 answer:
Elan Coil [88]2 years ago
5 0

Answer:

a=0.284\ m/s^2

Explanation:

Given that,

Initially, the spaceship was at rest, u = 0

Final velocity of the spaceship, v = 11 m/s

Distance accelerated by the spaceship, d = 213 m

We need to find the acceleration experienced by the occupants of the spaceship during the launch. It is a concept based on the equation of kinematics. Using the third equation of motion to find acceleration.

v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(11)^2-(0)^2}{2\times 213}\\\\a=0.284\ m/s^2

So, the acceleration experienced by the occupants of the spaceship is 0.284\ m/s^2.

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Unlike acceleration and velocity, speed does not need to specify
frosja888 [35]

Unlike acceleration and velocity, speed does not need to specify the direction of motion. Speed is a scalar quality.

4 0
2 years ago
Turner's treadmill runs with a velocity of -1.3 m/s and speeds up at regular intervals during a half-hour workout. after 25 min,
Travka [436]

Given : Initial velocity = -1.3 m/s

            Final Velocity = -6.5 m/s.

            Time = 25 minutes.

To find : Average acceleration.

Solution: We are given units in meter/second (m/s).

So, we need to convert time 25 minutes in seconds.

1 minute = 60 seconds.

25 minutes = 60*25 = 1500 seconds.

Formula for average acceleration is given by,

\frac{Final \ velocity -Initial \ velocity}{Final \ time - Initial \ time }

We are not given intial time, so we can take initial time =0.

Plugging values in the above formula.

Average \ acceleration = \frac{-6.5 -(-1.3)}{1500-0}

= \frac{-5.2}{1500}

= -0.003467

or Average \ acceleration = -3.467 \times 10^{-3}\ m/s^2..


3 0
2 years ago
Read 2 more answers
If you add 700 kJ of heat to 700 g of water at 70 degrees C, how much water is left in the container? The latent heat of vaporiz
makkiz [27]

Answer:A

Explanation:Find attached picture file for details

3 0
2 years ago
The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the ob
givi [52]

Answer:

Charge, Q=1.56\times 10^{-8}\ C

Explanation:

It is given that,

Electric field strength, E = 180000 N/C

Distance from a small object, r = 2.8 cm = 0.028 m

Electric field at a point is given by :

E=\dfrac{kQ}{r^2}

Q is the charge on an object

Q=\dfrac{Er^2}{k}

Q=\dfrac{180000\ N/C\times (0.028\ m)^2}{9\times 10^9\ Nm^2/C^2}

Q=1.56\times 10^{-8}\ C

So, the charge on the object is 1.56\times 10^{-8}\ C. Hence, this is the required solution.

7 0
2 years ago
A father demonstrates projectile motion to his children by placing a pea on his fork's handle and rapidly depressing the curved
MariettaO [177]

Answer:

4.17 m/s

Explanation:

To solve this problem, let's start by analyzing the vertical motion of the pea.

The initial vertical velocity of the pea is

u_y = u sin \theta = (7.39)(sin 69.0^{\circ})=6.90 m/s

Now we can solve the problem by applying the suvat equation:

v_y^2-u_y^2=2as

where

v_y is the vertical velocity when the pea hits the ceiling

a=g=-9.8 m/s^2 is the acceleration of gravity

s = 1.90 is the distance from the ceiling

Solving for v_y,

v_y = \sqrt{u_y^2+2as}=\sqrt{(6.90)^2+2(-9.8)(1.90)}=3.22 m/s

Instead, the horizontal velocity remains constant during the whole motion, and it is given by

v_x = u cos \theta = (7.39)(cos 69.0^{\circ})=2.65 m/s

Therefore, the speed of the pea when it hits the ceiling is

v=\sqrt{v_x^2+v_y^2}=\sqrt{2.65^2+3.22^2}=4.17 m/s

5 0
2 years ago
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