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Semenov [28]
3 years ago
15

A car heading north collides at an intersection with a truck heading east. if they lock together and travel 28 m.s at 46 degrees

north of east just after the collision, how fast was the car initially traveling? assume that the two vehicles have the same mass.
A) 30 m/s
B) 80 m/s
C) 20 m/s
D) 40 m/s
Physics
1 answer:
Dominik [7]3 years ago
4 0

Answer:

C) 20 m/s

Explanation:

The collision of the car and truck will give a single resultant velocity due to equal masses.

We'll treat this question as a vector question.

The velocity profile will form a right angle triangle.

To calculate the velocity of the car

Let the velocity be x

Sina = x/28

Sin46= x/28

0.7193*28 = x

20 = x

X = 20 m/s

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Which of the following is an instantaneous speed?
Brilliant_brown [7]

Answer:

A: All of the above

Explanation:

The instantaneous speed of an object is simply the current seed of the object at any given time. The SI unit is m/S and it is a vector quantity.

Therefore, according to the given options, they all have SI units that are consistent with distance and time which makes them all an example of instantaneous speed.

5 0
3 years ago
78 grams of a radioactive nuclei X undergoes radioactive decay. The half-life of X is 4.7 minutes. After 16.5 minutes, the remai
kipiarov [429]

<u>Answer:</u> The remaining sample of X is 6.9 grams.

<u>Explanation:</u>

All the radioactive reactions follow first order kinetics.

The equation used to calculate rate constant from given half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

We are given:

t_{1/2}=4.7min

Putting values in above equation, we get:

k=\frac{0.693}{4.7min}=0.147min^{-1}

The equation used to calculate time period follows:

N=N_o\times e^{-k\times t}

where,

N_o = initial mass of sample X = 78 g

N = remaining mass of sample X = ? g

t = time = 16.5 min

k = rate constant = 0.147min^{-1}

Putting values in above equation, we get:

N=78\times e^{-(0.147min^{-1}\times 16.5min)}\\\\N=6.9g

Hence, the remaining amount of sample X is 6.9 g

4 0
4 years ago
Currents during lightning strikes can be up to 50000 A (or more!). We can model such a strike as a 49500 A vertical current perp
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Answer:

59.4 N

Explanation:

The force exerted on a current-carrying wire due to a magnetic field perpendicular to the wire is given by

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where

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L is the length of the wire

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I = 49,500 A is the current

L = 1 m is the length (we want to find the force per each meter of length)

B=12 G = 12\cdot 10^{-4} T is the strength of the magnetic field

Therefore, the force on each meter of the current due to the magnetic field is:

F=(49,500)(1)(12\cdot 10^{-4})=59.4 N

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3 years ago
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