<span>The bullfrog is sitting at rest on the log. The force of gravity pulls down on the bullfrog. We can find the weight of the bullfrog due to the force of gravity.
weight = mg = (0.59 kg) x (9.80 m/s^2)
weight = 5.782 N
The bullfrog is pressing down on the log with a force of 5.782 newtons. Newton's third law tells us that the log must be pushing up on the bullfrog with a force of the same magnitude. Therefore, the normal force of the log on the bullfrog is 5.782 N</span>
I am going to need a picture for this question
Atmospheric electricity and storms, electrostatic control filters, and industrial electrostatic seperation as well as spark discharge. these are just a few. hope it helps.
Answer: The question has some missing details. The initial velocity given as u = -6.5i + 17j + 13k and the final velocity v = -2.8i + 17j -9.3k.
a) = (1.82i - 9.69k)m/s2
b) magnitude = 9.85m/s2
c) direction = 280.64 degree
Explanation:
The detailed and step is shown in the attachment.
<h2>
Answer:</h2>
The rate of deceleration is -0.14
<h2>
Explanation:</h2>
Using one of the equations of motion;
v = u + at
where;
v = final velocity of the boat = 0m/s (since the boat decelerates to a stop)
u = initial velocity of the boat = 25m/s
a = acceleration of the boat
t = time taken for the boat to accelerate/decelerate from u to v = 3 minutes
<em>Convert the time t = 3 minutes to seconds;</em>
=> 3 minutes = 3 x 60 seconds = 180seconds.
<em>Substitute the values of v, u, t into the equation above. We have;</em>
v = u + at
=> 0 = 25 + a(180)
=> 0 = 25 + 180a
<em>Make a the subject of the formula;</em>
=> 180a = 0 - 25
=> 180a = -25
=> a = -25/180
=> a = -0.14
The negative value of a shows that the boat is decelerating.
Therefore, the rate of deceleration of the speed boat is 0.14