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OLga [1]
2 years ago
9

A 2300 kg truck has put its front bumper against the rear bumper of a 2500 kg suv to give it a push. with the engine at full pow

er and good tires on good pavement, the maximum forward force on the truck is 18,000 n.
Physics
2 answers:
devlian [24]2 years ago
8 0

Answer:

Acceleration = a = 3.75 m/s^2

Explanation:

Mass of truck1 = m1 = 2300 kg

Mass of truck2 = m2 = 2500 kg  

Total mass = m = m1 + m2 = 2300 + 2500 = 4800 kg

Force exerted by truck1 = F = 18000 N

As both trucks are joint together so, behaving as single object. The acceleration can be found by Newton’s second law of motion.

F = ma  

a = F/m = 18000/4800

a = 3.75 m/s^2

valentina_108 [34]2 years ago
7 0
F=ma
m=total mass = 2300kg+2500kg=4800
F=18000N
a=?
a=F/m
a=18000/4800
a=3.8m/s^2
Final answer
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Car with a mass of 1210 kg moving at a velocity of 51 m/s.
2. What velocity must a 1340 kg car have in order to have the same momentum as a 2680 kg truck traveling at a velocity of 15 m/s to the west? 3.0 X 10^1 m/s to the west.

Hope i helped
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6 0
2 years ago
An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass
solniwko [45]

Answer:

A) The new amplitude = 0.048 m

B) Period T = 0.6 seconds

Explanation: Please find the attached files for the solution

4 0
2 years ago
Suppose you are designing an amplifier and loudspeaker system to use at a rock concert. You want to make it as loud as possible.
OverLord2011 [107]

Answer is given below

Explanation:

  • Audio power amplifiers are found in all types of sound systems, including sound reinforcement, public address and home audio systems, as well as musical instrument amplifiers such as guitar amplifiers.
  • This is the last electronic step in the general audio playback series before sending the signal to the loudspeaker.  So when we want maximum volume or loud sound, we have to get it with maximum output and high input and low output impedance
4 0
2 years ago
A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp
Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}

1. Finding Ф from the initial conditions:

-0.02 = 0.02cos(0 + \phi) => \phi = \pi

2. Finding time t at position y = 1 cm:

0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}

3. Find velocity v at time t from equation 2:

v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))

5 0
1 year ago
Read 2 more answers
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.75 A out of the jun
AlexFokin [52]

Answer:

number of electrons = 2.18*10^18 e

Explanation:

In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.

Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:

i_1+i_3=i_2                 (1)

i1 = 0.40 A

i2 = 0.75 A

you solve the equation i3 from the equation (1):

i_3=i_2-i_1=0.75A-0.40A=0.35A

Next, you take into account that 1A = 1C/s = 6.24*10^18

Then, you have:

0.35A=0.35\frac{C}{s}=0.35*\frac{6.24*10^{18}e}{s}=2.18*10^{18}\frac{e}{s}

The number of electrons that trough the wire 3 is 2.18*10^18 e/s

3 0
2 years ago
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