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antiseptic1488 [7]

3 years ago

10

Determine the field strength, E, experienced by a test charge, q, if a charge of 7.0 × 10-5 coulombs is placed on q and a force

of 5.2 N is applied.

2 answers:

kkurt [141]3 years ago

7 0

Formula for feild strength= F/q
q=7.0^10-5 coulombs
F=5.2 N
E=5.2 / 7.0^10-5
E=

Evgesh-ka [11]3 years ago

5 0

E = F/q

= 5.2 N / (7* 10⁻⁵) C

= 7.429 * 10⁴ N/C

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A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on

MAXImum [283]

Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

$Ec=\frac{1}{2} *m*v^{2}$

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

$W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}$

$W=\frac{1}{2} *m*(v2^{2}-v1^{2})$

In this case:

W=?
m= 2,145 kg
v2= 12 $\frac{m}{s}$

v1= 25 $\frac{m}{s}$

Replacing:

$W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})$

W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

3 0

3 years ago

Can someone help me with these

Lostsunrise [7]

Answer:

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Explanation:

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8 0

3 years ago

6) Bill and Amy want to ride their bikes from their neighborhood to school which is 14.4

Studentka2010 [4]

Answer:

Amy's speed is 2/3 faster than Bill's

Explanation:

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3 0

3 years ago

Consider that a clay model of a tiger has a mass of 0.195 kg and glides on ice at a speed of 0.75 m/s. It hits another clay mode

juin [17]

Answer:

The final velocity after the collision is 0.27 m/s.

Explanation:

Given that,

Mass of tiger, m = 0.195 kg

Initial speed of tiger model, v = 0.75 m/s

Mass of another clay model, m' = 0.335 kg

Initially, second model is at rest, v' = 0

We need to find the final velocity after the collision. It is a case of inelastic collision. Using the conservation of linear momentum as :

$mv+m'v'=(m+m')V\\\\V=\dfrac{mv+m'v'}{(m+m')}\\\\V=\dfrac{0.195\times 0.75}{(0.195 +0.335 )}\\\\V=0.27\ m/s$

So, the final velocity after the collision is 0.27 m/s.

6 0

4 years ago

The outside diameter of your teacher's rear bicycle tire is 16 inches. How far will he travel if the rear wheel makes 1200 revol

FinnZ [79.3K]

Answer:

241,274.32 inches

Explanation:

How far will he travel if the rear wheel makes 1200 revolutions on the road?

Since the rear wheel makes one revolution in the distance of a circumference of a circle, C with diameter, d = 16 inches

C = πd²/4

So, the distance, travelled in 1200 revolutions is D = 1200 × C = 1200πd²/4

Substituting d = 16 into D, we have

D = 1200πd²/4

D = 1200π(16)²/4

D = 76800π

D = 241,274.32 inches

3 0

3 years ago