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mart [117]
2 years ago
8

An ice skater slides in a straight line across the ice. As she passes a certain point, her velocity is 24 km/hr. Two seconds lat

er it is 13 km/hr. What is the acceleration of the skater?
5.5 km/hr/sec
-5.5 km/hr/sec
11 km/hr/sec
-11 km/hr/sec
Physics
1 answer:
gulaghasi [49]2 years ago
5 0
Acceleration = (change in speed) / (time for the change)

Change in speed = (later speed) - (earlier speed) = (13 - 24) = -11 km/hr
Time for the change = 2 seconds

Acceleration = (-11 km/hr) / (2 sec)

Acceleration = -5.5 km/hr-sec  (B)
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An electron moves with velocity v⃗ =(5.8i−6.7j)×104m/s in a magnetic field B⃗ =(−0.81i+0.60j)T.
Minchanka [31]

Answer:

Fₓ = 0,  F_{y} = 0  and  F_{z}<em> = - 3.115 10⁻¹⁵   N</em>

Explanation:

The magnetic force given by the expression

       F = q v xB

the bold are vectors,  the easiest analytical way to determine this force in solving the determinant

   F = q \left[\begin{array}{ccc}i&j&k\\5.8&-6.7&0\\-0.81&0.6&0\end{array}\right]  10^{4}

   F = 1.6 10⁻¹⁵ [ i( 0-0) + j (0-0) + k^( 5.8 0.60 - 0.81 67) ]

   F =i^0 + j^0   - k^  3.115 10⁻¹⁵   N

   

Fₓ = 0

F_{y} = 0

F_{z}<em> = - 3.115 10⁻¹⁵   N</em>

6 0
3 years ago
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Using Figure 2, what is the momentum of Train Car A before the collision?
Bess [88]

Answer:

Option A. 180000 Kgm/s.

Explanation:

From the question given above, the following data were obtained:

For Train Car A:

Mass of train car A = 45000 Kg

Velocity of train car A = 4 m/s

Momentum of train car A =?

For Train Car B:

Mass of train car B = 45000 Kg

Velocity of train car B = 0 m/s

Momentum is simply defined as the product of mass and velocity. Mathematically, it can be expressed as:

Momentum = mass × velocity

With the above formula, the momentum of train car A before collision can be obtained as follow:

Mass of train car A = 45000 Kg

Velocity of train car A = 4 m/s

Momentum of train car A =?

Momentum = mass × velocity

Momentum = 45000 × 4

Momentum of train car A = 180000 Kgm/s

5 0
2 years ago
The traffic on the freeway is moving at a constant speed of 24 m/sm/s. What distance does the traffic travel while the car is mo
ExtremeBDS [4]

Incomplete question as there is so much information is missing.The complete question is here

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

Answer:

Distance traveled=240 m

Explanation:

Given data

Initial velocity of car v₀=0 m/s

Final velocity of car vf=24 m/s

Distance traveled by car S=120 m

To find

Distance does the traffic travel

Solution

To find the distance first we need to find time, for time first we need acceleration

So

(V_{f})^{2}=(V_{o})^{2}+2aS\\  So\\a=\frac{(V_{f})^{2}-(V_{o})^{2} }{2S}\\ a=\frac{(24m/s)^{2}-(0m/s)^{2} }{2(120)}\\a=2.4 m/s^{2}

As we find acceleration.Now we need to find time

So

V_{f}=V_{i}+at\\t=\frac{V_{f}-V_{i}}{a}\\t=\frac{(24m/s)-(0m/s)}{(2.4m/s^{2} )}\\t=10s

Now for distance

So

Distance=velocity*time\\Distance=(24m/s)*(10s)\\Distance=240m

7 0
3 years ago
Someone please help me!
beks73 [17]

Answer:

current = 8,750

Explanation:

you will need to use the formula i = v/r to get this answer

3 0
3 years ago
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