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GenaCL600 [577]
3 years ago
6

How many miles is in 7 blocks?

Physics
1 answer:
borishaifa [10]3 years ago
8 0

Answer:

Different street blocks are different lengths, so it won't be possible to answer this.

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Please help with vectors (will give BRAINLIEST answer)
zhenek [66]

just analyze it in this way:

20cos30*=10( radical 3 )

20sin30*=10

7 0
3 years ago
Which sequence correctly reflects the order in which electrical charges occur during an action potential, from first to last?
Lady_Fox [76]
This sequence refer to how neurons send messages electrochemically. When the neurons are at rest, they are at its resting potential at -70 millivolts. If the neurons in the brain send messages, a spike would occur until it reaches the threshold potential at -55 mV. If it reaches the threshold potential, then an action potential occurs.Hence, the sequence would be letter C.
4 0
3 years ago
A diamond is underwater; A light ray enters one face of the diamond, then travels at an angle of 30 degrees with respect to the
otez555 [7]
According to snells law 

<span>n1 sin theta1 = n2 sin theta2
</span>n1  = 1.333 (water)
<span>n2  = 2.42 (diamond)</span>
 it is given that theta =30 degrees so 
by putting the values we have 

<span>1.333 sin theta = 2.42 sin 30 </span>

<span>sin theta = (2.42/1.333) *0.5 =65.2 degree
</span>so our conclusion is 

<span>the ray's angle of incidence θ1 on the diamond</span> = 65.2 degree.
hope this helps
4 0
3 years ago
Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an
joja [24]

Answer:

Approximately 9.7\; \rm m \cdot s^{-2}.

Explanation:

Assuming that there is no other force on this vehicle, the 16000\; \rm N force from the road would be the only force on this vehicle. The net force would then be equal to this 16000\; \rm N\! force. The size of the net force would be 16000\; \rm N\!\!.

Let m denote the mass of this vehicle and let \Sigma F denote the net force on this vehicle.  

By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, a, would be:

\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}.

For this vehicle, \Sigma F = 16000\; \rm N whereas m = 1650\; \rm kg. The acceleration of this vehicle would be:

\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}.

8 0
3 years ago
MathPhys Help pls Tysm
HACTEHA [7]

Answer:

8.75

Explanation:

First, find the force of friction.

Kinetic energy = work done by friction

½ mv² = Fd

½ (3.9 kg) (2.9 m/s)² = F (1.4 m)

F = 11.7 N

Next, find the distance at the new velocity.

Kinetic energy = work done by friction

½ mv² = Fd

½ (3.9 kg) (2.5 × 2.9 m/s)² = (11.7 N) d

d = 8.75 m

3 0
3 years ago
Read 2 more answers
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