Your teacher's SUV can go from rest to 25 m/s (roughly 55mph) in 10 seconds, The car's velocity changes at a uniform rate.

A leaf fell from a tree branch. The path it followed is shown in the diagram below.

prisoha [69]

The leaf fell at the crooked path instead of straight down because air currents and gravity applied changing and unbalanced forces to the leaf.

<h3>What is an air current?</h3>

An air current is defined as the changes in atmospheric pressure that causes the movement of air from one area to another.

When a leaf is detached naturally from the tree, it won't fall straight down to the floor but will fall a distance away from the tree due to the action of air current and some unbalanced forces.

Learn more about leaf here:

brainly.com/question/24234175

#SPJ1

4 0

2 years ago

A speedboat is approaching a dock at 25 m/s (56 mph). When the dock is 150 m away, the driver begins to slow down. a) What accel

trasher [3.6K]

Answer:

a) -2.038 m/s²

b) 40.33 mph

c) 312.5 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-25^2}{2\times 150}\\\Rightarrow a=-2.083\ m/s^2$

Acceleration of the boat is -2.083 m/s² if the boat will stop at 150 m.

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -1\times 150+25^2}\\\Rightarrow v=18.03\ m/s$

Speed of the boat by when it will hit the dock is 18.03 m/s

Converting to mph

$1\ mile=1609.34\ m$

$1\ h=3600\ seconds$

$18.03\times \frac{3600}{1609.34}=40.33\ mph$

Speed of the boat by when it will hit the dock is 40.33 mph

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-25^2}{2\times -1}\\\Rightarrow s=312.5\ m$

The distance at which the boat will have to start decelerating is 312.5 m

5 0

3 years ago

Calculate the fraction of methyl isonitrile molecules that have an energy of 160.0 kJ or greater at 510 K .

soldi70 [24.7K]

Answer:

$4.0932672025\times 10^{-17}$

Explanation:

$E_a$ = Activation energy = 160 kJ

T = Temperature = 510 K

R = Universal gas constant = 8.314 J/mol K

The fraction of energy is given by

$f=e^{-\dfrac{E_a}{RT}}\\\Rightarrow f=e^{-\dfrac{160000}{8.314\times 510}}\\\Rightarrow f=4.0932672025\times 10^{-17}$

The fraction of energy is $4.0932672025\times 10^{-17}$

3 0

3 years ago

Two balloons are charged with an identical quantity and type of charge: -0.0025 C. They are held apart at a separation distance

jok3333 [9.3K]

Answer:

F = 878.9 N

Explanation:

The electrostatic force of attraction or repulsion is given by Coulomb's Law as follows:

F = kq₁q₂/r²

where,

F = Force pf repulsion between balloons = ?

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = q₂ = magnitudes of 1st and 2nd charge = 0.0025 C

r = distance between balloons = 8 m

Therefore,

F = (9 x 10⁹ N.m²/C²)(0.0025 C)(0.0025 C)/(8 m)²

3 0

3 years ago

A 0.5 kg air-track car is attached to the end of a horizontal spring of constant k = 20 N/m. The car is displaced 15 cm from its

Dimas [21]

Find the amount of work that the spring does. This can be found using the equation 1/2kx^2. Then, you must set that equal to the amount of kinetic energy the car has. This is possible thanks to the work-energy theorem.

1/2kx^2 = 1/2mv^2

Solve to find velocity. Remember, the spring is displaced .15 m, not 15!

To find the acceleration, use F = ma. The force being applied to the car is kx, and you know the mass. You do the math.

For problem C I don't know, haven't done that yet in my class. Sorry!

3 0

3 years ago