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Alex73 [517]
3 years ago
5

a charge of 30. coulombs passes through a 24-ohm resistor in 6.0 seconds. what is the current through the resistor? (1) 1.3 a (3

) 7.5 a (2) 5.0 a (4) 4.0 a
Physics
2 answers:
DochEvi [55]3 years ago
7 0
As I = Q/t (charge/time) the answer is 5a.

30/6 = 5
uranmaximum [27]3 years ago
3 0
Current is the rate of flow of charges.

I = Q / t

Where I is current in Ampere,  Q is charges in Coulombs, t is time in seconds.

I = 30 / 6

I = 5 A.

Option (2) is the answer.
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Suppose a treadmill has an average acceleration of 4.7x10^-3 m/s. a)how much does its speed change after 5min? b)if the treadmil
Mars2501 [29]

Answer:

a)Change in the speed  = 1.41 m/s

b)The final speed will be 3.11 m/s

Explanation:

Given that

Acceleration ,a= 4.7 x 10⁻³ m/s²

a)

We know that

v= u + a t

v=final speed ,u=initial speed

t= time ,a= acceleration

Change in the speed

v- u = a t

t= 5 min  = 5 x 60 s = 300 s

v- u = 4.7 x 10⁻³ x 5 x 60 m/s

v-u = 1.41 m/s

Change in the speed  = 1.41 m/s

b)

Given that

u= 1.7 m/s

v-u = 1.41 m/s

v= 1.7 + 1.41 m/s

v=3.11 m/s

The final speed will be 3.11 m/s

8 0
3 years ago
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

5 0
3 years ago
A 565 N rightward force pulls a large box across the floor with a constant velocity of 0.75 m/s. If the coefficient of friction
alukav5142 [94]
The first thing you should know is that the friction force is equal to the coefficient of friction due to normal force.
 Therefore, clearing the normal force we have:
 The friction is 565N.
 (565 / 0.8) = 706.25N. weight.
6 0
3 years ago
Question 7
nikdorinn [45]

Answer:

D

Explanation:

im not sure just sounds right

7 0
3 years ago
A 2500 N force accelerates a car at a rate of 3.0 m/s^2. What is the car’s mass? 250 kg
Ronch [10]

Apply Newton's second law to the car's motion:

F = ma

F = net force, m = mass, a = acceleration

Given values:

F = 2500N, a = 3.0m/s²

Plug in and solve for m:

2500 = m(3.0)

m = 830kg

5 0
3 years ago
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