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scZoUnD [109]
3 years ago
11

A large truck collides with a small car. True or False: The truck exerted a greater magnitude force on the car than the car exer

ted on the truck. True False
Physics
1 answer:
Reptile [31]3 years ago
5 0

Answer:

False.

Explanation:

The forces on the car and truck are equal and opposite. The equal forces cause accelerations of the truck and car inversely proportional to their mass. That is, If the Truck A exerts a force FAB on car B, then the car will exert a force FBA on the truck. Therefore,

FBA = −FAB

However, this can be explained by Newton's second law. Let's say the truck has mass M and the car has mass m. If the magnitude of the force that both vehicles experience is F, then the magnitudes of their respective accelerations are:

atruck = F/M

acar = F/m

and combining these we get:

atruck/acar = m/M

So if the mass of the car is a lot less than the mass of the truck, then the acceleration of the truck is much smaller than the acceleration of the car, and if you were to watch the collision, the truck would pretty much seem like it's motion was unaffected, but the car's motion will change quite a bit.

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Ethylene glycol, the primary ingredient in antifreeze, has the chemical formula C2H6O2. The radiator fluid used in most cars is
disa [49]

Answer:

-30 °C

Explanation:

First, we have to calculate the molality (m) of the solution. If the solution is 50% C₂H₆O₂ by mass. It means that in 100 g of solution, the are 50 g of solute (C₂H₆O₂) and 50 g of solvent (water).

The molar mass of C₂H₆O₂ is 62.07 g/mol. The moles of solute are:

50 g × (1 mol / 62.07 g) = 0.81 mol

The mass of the solvent is 50 g = 0.050 kg.

The molality is:

m = 0.81 mol / 0.050 kg = 16 m

The freezing-point depression (ΔT) can be calculated using the following expression.

ΔT = Kf × m = (1.86 °C/m) × 16 m = 30 °C

where,

Kf: freezing-point constant

The normal freezing point for water is 0°C. The freezing point of the radiator fluid is:

0°C - 30°C = -30 °C

8 0
3 years ago
A scooter has wheels with a diameter of 120 mm. What is the angular speed of the wheels when the scooter is moving forward at 6.
nirvana33 [79]

To develop this problem we will apply the concepts related to angular kinematic movement, related to linear kinematic movement. Linear velocity can be described in terms of angular velocity as shown below,

v = r\omega \rightarrow \omega = \frac{v}{r}

Here,

v = Lineal velocity

\omega= Angular velocity

r = Radius

Our values are

v = 6/ms

r = \frac{d}{2} = \frac{120*10^{-3}}{2} = 0.06m

Replacing to find the angular velocity we have,

\omega = \frac{6m/s}{0.06m}

\omega = 100rad/s

Convert the units to RPM we have that

\omega = 100rad/s (\frac{1rev}{2\pi rad})(\frac{60s}{1m})

\omega = 955.41rpm

Therefore the angular speed of the wheels when the scooter is moving forward at 6.00 m/s is 955.41rpm

4 0
3 years ago
The area under acceleration time garph represents?​
Keith_Richards [23]

Answer:

Change in Velocity because

at = v

Explanation:

Remeber area is length times Width. In this case, the area under a accleraton vs time graph is Accleration Times Time. Which is velocity

8 0
2 years ago
Read 2 more answers
A blue train of mass 50 kg moves at 4 m/s toward a green train of 30 kg initially at rest. What is the initial momentum of the g
soldi70 [24.7K]

Answer:

I think answer is zero

bcz momentum=mass×velocity

body was initially at rest it means its velocity is zero

30×0=0

6 0
3 years ago
What is malware short for?
yaroslaw [1]
The correct answer is C
8 0
3 years ago
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