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Salsk061 [2.6K]
3 years ago
6

Scientists develop theories to explain natural events. What causes scientists to reject even long-held theories and replace them

with newer theories?
Physics
1 answer:
cricket20 [7]3 years ago
8 0
New discoveries and proof of new information
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A 1.5 kilogram cart initially moves at 2.0 meters per second. It is brought to rest by a constant net forcé in 3.0 second. What
lubasha [3.4K]

The net force on the cart is

                   Force = (mass) x (acceleration) .

We know the mass = 1.5 kilogram, but we need to
calculate the acceleration.

               Acceleration = (change in speed) / (time for the change)

Change in speed = (ending speed) minus (beginning speed)

                               =    ( 0 )  -  ( 2 m/s )  =   -2 m/s

Acceleration =  (change in speed) / (time for the change)

                       =       ( -2 m/s )   /   (3 sec)

                       =           - 2/3  m/s² .

Finally ...        Net force = (mass) x (acceleration)

                                       =  (1.5 kilogram) x ( - 2/3 m/s²)

                                       =  ( - 3/2 x 2/3 )  (kg-m/s²)

                                       =       - 1  newton .

The negative sign on the force means the force is applied
opposite to the direction the cart is moving.  That gives
negative acceleration (slowing down), and the cart
eventually stops.   

3 0
4 years ago
What is the current in a wire if 3.4 x 10^19 electrons pass by a point in this wire every 60 seconds?
Misha Larkins [42]
The current is defined as the amount of charge Q that passes through a certain point of a wire in a time \Delta t:
I= \frac{Q}{\Delta t}

But the charge Q flowing in the wire is just the charge of a single electron, e, times the number of electrons N:
Q=Ne
so the current can be rewritten as
I= \frac{Ne}{\Delta t}

Using N=3.4 \cdot 10^{19}, e= 1.6 \cdot 10^{-19} C (charge of one electron), and \Delta t = 60 s, we find the current:
I= \frac{Ne}{\Delta t}= \frac{(3.4 \cdot 10^{19})(1.6 \cdot 10^{-19}C}{60 s}=9.1 \cdot 10^{-2}A

Therefore correct answer is A).
8 0
3 years ago
.It takes time 8 min 20 seconds for light to reach from sun to the earth surface.If speed of light is taken to be 3times10^(5)km
erma4kov [3.2K]

Answer:

hjjiisisijjwjj

Explanation:

jjsjqiqu##uwuuw

3 0
3 years ago
A 1.70 m cylindrical rod of diameter 0.500 cm is connected to a power supply that maintains a constant potential difference of 1
Natalija [7]

Answer:

(1) The resistivity of the rod at 20 °C is 8.652×10^-6 ohm-meter.

(2) The temperature coefficient of resistivity at 20 °C is 0.00125/°C

Explanation:

(1) Resistance at 20 °C = V/I = 14/18.7 = 0.749 ohm

Length = 1.7 m

Diameter (d) = 0.5 cm = 0.5/100 = 0.005 m

Area = πd^2/4 = 3.142×0.005^2/4 = 1.96375×10^-5 m^2

Resistivity at 20 °C = resistance × area/length = 0.749×1.96375×10^-5/1.7 = 8.652×10^-6 ohm-meter.

(2) Resistance at 92 °C = V/I = 14/17.2 = 0.814 ohm

Temperature coefficient at 20 °C = (0.814/0.749 - 1) ÷ (92 - 20) = (1.09 -1) ÷ 72 = 0.09 ÷ 72 = 0.00125/°C

5 0
4 years ago
Read 2 more answers
A spring toy jumps up from the floor and comes back down to the floor. Its initial speed is 12 m/s. What is the
dybincka [34]

Answer:

0\; \rm m \cdot s^{-1}.

Explanation:

In a projectile motion, the kinetic energy (KE) of the projectile is converted to gravitational potential energy (GPE) and then from GPE back to KE.

In this example, the spring toy is the projectile. Refer to the diagram attached.

  • The spring toy started with a velocity of 12\; \rm m\cdot s^{-1}, meaning that its initial KE is non-zero.
  • On the way up, the KE of this spring toy gets converted to GPE.
  • At the top of the trajectory, the GPE of this toy is maximized while its KE is minimized (zero).
  • As the toy returns to the ground, the GPE of this toy gets converted back to KE.
  • The GPE of this toy is zero when the toy is on the ground.

In other words, the kinetic energy (KE) of this toy would be 0 when it is at the top of the trajectory and GPE is maximized.

Since the KE of this toy at the top of the trajectory would be 0, the velocity of this toy at that moment would also be 0\!.

8 0
3 years ago
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