Thermals created by warm air rising and cold air sinking are called__

A student conducts an experiment in which a cart is pulled by a variable applied force during a 2 s time interval. In trial 1, t

fredd [130]

Answer:

The answer is "Including all three studies of 0s to 2s, that shift in momentum is equal".

Explanation:

Its shift in momentum doesn't really depend on the magnitude of its cars since the forces or time are similar throughout all vehicles.

Let's look at the speed of the car

$F = m a\\\\a =\frac{F}{m}$

We use movies and find lips

$\to v = v_0 + a t\\\\\to v = v_0 + (\frac{F}{m}) t$

The moment is defined by

$\to p = m v$

The moment change

$\Delta p = m v - m v_0$

Let's replace the speeds in this equation

$\Delta p = m (v_0 + \frac{F}{m t}) - m v_0\\\\\Delta p = m v_0 + F t - m v_0\\\\\Delta p = F t$

They see that shift is not directly proportional to the mass of cars since the force and time were the same across all cars.

5 0

3 years ago

A 65.0 kg ice skater standing on frictionless ice throws a 0.15 kg snowball horizontally at a speed of 32.0 m/s. What is the vel

spin [16.1K]

Answer:

(d) 0.07 m/s

Explanation:

Given Data

Snowball mass m₁=0.15 kg

Ice skater mass m₂=65.0 kg

Snowball velocity v₁=32.0 m/s

To find

Velocity of Skater v₂=?

Solution

From law of conservation of momentum

$m_{1}v_{1}=m_{2}v_{2}\\ v_{2}=\frac{m_{1}v_{1}}{m_{2}}\\ v_{2}=\frac{(0.15kg)(32.0m/s)}{65.0kg}\\ v_{2}=0.0738m/s\\or\\v_{2}=0.07 m/s$

So Option d is correct one

5 0

3 years ago

A branch falls from a tree how fast is the branch moving after drug .28 seconds

alexdok [17]

Acceleration=9.81m/s^2
initial velocity=0m/s
time=.28s

We have to find final velocity.

The equation we use is

Final velocity=initial velocity+acceleration x time

Vf=0m/s+(9.81m/s^2)(.28s)
Vf=2.7468m/s

We would round this to:

Vf (final velocity)=2.7m/s

6 0

3 years ago

A hockey puck has a coefficient of kinetic friction of μk = .35. If the puck feels a normal force (FN) of 5 N, what is the frict

alina1380 [7]

Answer:

The frictional force is $F_f = 1.75 \ N$

Explanation:

From the question we are told that

The coefficient of kinetic force is μk = 0.35

The normal force felt by the puck is $F_N = 5 \ N$

Generally the frictional force that acts on the puck is mathematically represented as

$F_f = \mu_k * F_N$

=> $F_f = 0.35 * 5$

=> $F_f = 1.75 \ N$

3 0

3 years ago

An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu

ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, $qE \times S + mg \times S = 0.5 \times mv^{2}$

$1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}$

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight and the above equation will be as follows.

$(1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}$

v = $sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}$

= 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

8 0

3 years ago

Thermals created by warm air rising and cold air sinking are called___?