Answer:
Explanation:
a=v-u/t
a=acceleration
v=final velocity
u=initial velocity
t=tme taken
we need to convert from kph to ms⁻¹
v= 150*1000/60*60= 41.67ms⁻¹
u= 120*1000/60*60= 33.33ms⁻¹
t= 2*60= 120s
a=41.67-33.33/120
a=8.34/120
a=0.0694ms⁻²
Answer:
v = 2.18m/s
Explanation:
In order to calculate the speed of Betty and her dog you take into account the law of momentum conservation. The total momentum before Betty catches her dog must be equal to the total momentum after.
Then you have:
(1)
M: mass Betty = 40kg
m: mass of the dog = 15kg
v1o: initial speed of Betty = 3.0m/s
v2o: initial speed of the dog = 0 m/s
v: speed of both Betty and her dog = ?
You solve the equation (1) for v:

The speed fo both Betty and her dog is 2.18m/s
Answer:
<h2>42 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question
mass = 7 kg
acceleration = 6 m/s²
We have
force = 7 × 6 = 42
We have the final answer as
<h3>42 N</h3>
Hope this helps you
The third equation of free fall can be applied to determine the acceleration. So that Paola's acceleration during the flight is 39.80 m/
.
Acceleration is a quantity that has a direct relationship with velocity and also inversely proportional to the time taken. It is a vector quantity.
To determine Paola's acceleration, the third equation of free fall is appropriate.
i.e
=
± 2as
where: V is the final velocity, U is the initial velocity, a is the acceleration, and s is the distance covered.
From the given question, s = 20.1 cm (0.201 m), U = 4.0 m/s, V = 0.
So that since Poala flies against gravity, then we have:
=
- 2as
0 =
- 2(a x 0.201)
= 16 - 0.402a
0.402a = 16
a = 
= 39.801
a = 39.80 m/
Therefore Paola's acceleration is 39.80 m/
.
Visit: brainly.com/question/17493533
Answer:
time=4s
Explanation:
we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion
, where
max=
Given that, at i(2)=
⇒
⇒
⇒
Applying logarithm on both sides,
⇒
⇒
⇒
Now substitute 
⇒
⇒
⇒
Applying logarithm on both sides,
⇒
⇒
⇒
now subs. 
⇒
also 
⇒
⇒