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Romashka [77]
3 years ago
14

A car is traveling at 30 m/s ( approximately 67 mph) on I-95 highway. The driver applies the brakes and the car comes to a stop

in 4.0 s . What is the car's acceleration, in m/s2
Physics
1 answer:
o-na [289]3 years ago
7 0

Answer:

step 1 the initial velocity of the car is u = 30 m/s.the final velocity of the car is v = 0 m/s. the time taken by

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Convert 7.93 lbs into grams. Hint: 1 kg = 2.2 lbs
Alekssandra [29.7K]

Answer:

7.93 lbs is equal to 3596.987 grams.

Explanation:

The weight in grams is equal to the pounds multiplied by 453.59237.

So... you would multiply 7.93 by 453.59237.

7.93 x 453.59237 = 3596.987

Hope that helped!

6 0
3 years ago
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A 60 kg box is lifted by a rope a distance of 10 meters straight up at constant speed. how much power is required to complete th
Nadya [2.5K]
Power=Work/Time
The work done is the energy required to lift the box, fighting the force of gravity. So, Work=Potential energy of the box at 10 meters.

W=PE=mgh=(60)(9.8)(10)=5880J
Finally,
P=W/T=(5880)/(5)=1176Watt

So the answer is 1176 Watts
3 0
3 years ago
Read 2 more answers
How could a slow moving train had the same momentum as a high-speed bullet
Flura [38]

Answer:

it is sooo easy u need to use magnetic panels on the rail and put the magnet on the train it works under the principal of magnetic

6 0
3 years ago
The component of the external magnetic field along the central axis of a 78-turn circular coil of radius 34.0 cm decreases from
grigory [225]

Answer:

Induced current, I = 18.88 A

Explanation:

It is given that,

Number of turns, N = 78

Radius of the circular coil, r = 34 cm = 0.34 m

Magnetic field changes from 2.4 T to 0.4 T in 2 s.

Resistance of the coil, R = 1.5 ohms

We need to find the magnitude of the induced current in the coil. The induced emf is given by :

\epsilon=-N\dfrac{d\phi}{dt}

Where

\dfrac{d\phi}{dt} is the rate of change of magnetic flux,

And \phi=BA

\epsilon=-NA\dfrac{dB}{dt}

\epsilon=-78\times \pi (0.34)^2\dfrac{(0.4-2.4)}{2}

\epsilon=28.32\ V

Using Ohm's law, \epsilon=I\times R

Induced current, I=\dfrac{\epsilon}{R}

I=\dfrac{28.32}{1.5}

I = 18.88 A

So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.

5 0
3 years ago
A 2.5 m -long wire carries a current of 8.0 A and is immersed within a uniform magnetic field B⃗ . When this wire lies along the
leva [86]

Answer:

Explanation:

Let the magnetic field be B = B₁i + B₂j + B₃k

Force = I ( L x B )  , I is current , L is length and B is magnetic field .

In the first case

force = - 2.3 j N

L = 2.5 i

puting the values in the equation above

- 2.3 j = 8 [ 2.5 i x ( B₁i + B₂j + B₃k )]

= - 20 B₃ j + 20 B₂ k

comparing LHS and RHS ,

20B₃ = 2.3

B₃ = .115

B₂ = 0

In the second case

L = 2.5 j

Force = I ( L x B )

2.3i−5.6k = 8 ( 2.5 j x (B₁i + B₂j + B₃k )

=  - 20 B₁ k + 20B₃ i

2.3i−5.6k = - 20 B₁ k + 20B₃ i

B₃ = .115

B₁ = .28

So magnetic field B = .28 i + .115 B₃

Part A

x component of B = .28 T

Part B

y component of B = 0

Part C

z component of B = .115 T .

8 0
3 years ago
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