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The 1350-kg car has a velocity of 24 km/h up the 8-percent grade when the driver applies more power for 18 s to bring the car up

Brrunno [24]

Answer:

Explanation:

We shall apply concept of impulse to solve the problem .

Impulse = force x time

impulse = change in momentum

force x time = change in momentum

initial speed u = 24 km/h = 6.67 m /s

final speed v = 65 km/h = 18.05 m /s

change in momentum = m v - mu

= m ( v-u )

= 1350 ( 18.05 - 6.67 )

= 15363 kg m/s

F x 18 = 15363

F = 853.5 N .

4 0

2 years ago

What units would you use to measure a red blood cell ? (nanometers micrometers millimeters

Stels [109]

Viruses are infectious particles made of nucleic acid (DNA or RNA) and protein. Viruses are 100 times smaller than the size of bacteria.

4 0

2 years ago

Read 2 more answers

"determine the resultant internal loadings acting at the cross sections at points f and g of the frame. set θ = 27º and t = 178

Leviafan [203]

Hi you didn't provide any images to solve the question, hence I am going to solve a different question of same concept so you can have an idea how to tackle such types of questions.(please refer to the attachment for question)

Answer:

Please refer to the attachment for answers and explanation

Explanation:

Please refer to the attachment for answers and explanation

5 0

2 years ago

A hockey player applies an average force of 80.0 N to a

denpristay [2]

Answer:

$Impulse = 8.0Ns$

Explanation:

Given

$Force = 80.0N$

$Mass = 0.25kg$

$Time = 0.10s$

Required

Determine the impulse

The impulse is calculated as follows:

$Impulse = Force * Time$

Substitute values for Force and Time

$Impulse = 80.0N * 0.10s$

$Impulse = 8.0Ns$

Hence, the impulse experienced is 8.0Ns

5 0

2 years ago

The quantity of charge passing through a surface of area 1.82 cm2 varies with time as q = q1 t 3 + q2 t + q3 , where q1 = 5.2 C/

Brrunno [24]

Answer:

Current through the surface at t = 1.1 s is 21.37 A.

Explanation:

The charge is passing through a surface of area varies with time as :

$q=q_1t^3+q_2t+q_3$

Here,

$q_1=5.2\ C/s^3\\\\q_2=2.5\ C/s\\\\q_3=6.5\ C$

t is in seconds

$q=5.2t^3+2.5t+6.5$

The rate of change of electric charge is called electric current. It is given by :

$I=\dfrac{dq}{dt}\\\\I=\dfrac{d(5.2t^3+2.5t+6.5)}{dt}\\\\I=15.6t^2+2.5$

At t = 1.1 s, Current,

$I=15.6(1.1)^2+2.5\\\\I=21.37\ A$

So, the instantaneous current through the surface at t = 1.1 s is 21.37 A.

3 0

2 years ago