1) If I have 10 A of current running through my series circuit (which has 5

4. The earth exerts a gravitational force of 3.5 N on an object. What is the mass of

Romashka-Z-Leto [24]

Answer:

I'm not sure

Explanation:

sorry I'm not smart that's why I have this app

8 0

3 years ago

A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences

NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, $q=3\times 10^{-6}\ C$

Mass of charge particle, $m=2\times 10^{-6}\ C$

Speed, $v=5\times 10^{6}\ m/s$

Acceleration, $a=3\times 10^{4}\ m/s^2$

We need to find the minimum magnetic field that would produce such an acceleration. So,

$ma=qvB\ sin\theta$

For minimum magnetic field,

$ma=qvB$

$B=\dfrac{ma}{qv}$

$B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}$

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0

3 years ago

At a baseball game, the batter hit a fly ball at time t = 0 s. The outfielder caught the ball at t = 5.8 s. When was the ball at

Agata [3.3K]

We have the following equation for height:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
Where,
a: acceleration
vo: initial speed
h0: initial height.
The value of the acceleration is:
a = -g = -9.8 m / s ^ 2
For t = 0 we have:
h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
h (0) = h0
h0 = 0 (reference system equal to zero when the ball is hit).
For t = 5.8 we have:
h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
(1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
vo = (1/2) * (9.8) * (5.8)
vo = 28.42
Substituting values we have:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
Rewriting:
h (t) = -4.9 * t ^ 2 + 28.42 * t
The maximum height occurs when:
h '(t) = -9.8 * t + 28.42
-9.8 * t + 28.42 = 0
t = 28.42 / 9.8
t = 2.9 seconds.
Answer:
The ball was at maximum elevation when:
t = 2.9 seconds.

8 0

3 years ago

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A sample of an unknown substance has a mass of 0.465 kg. if 3,000.0 j of heat is required to heat the substance from 50.0°c to 1

hammer [34]

The specific heat of the substance will be 0.129 J/g°C.

<h3>What is specific heat capacity?</h3>

The amount of heat required to increase a substance's temperature by one degree Celsius is known as specific heat capacity.

Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.

The given data in the problem is;

Q is the amount of energy necessary to raise the temperature = 3,000.0 j

M is the mass= 0.465 kg.

Δt is the time it takes to raise the temperature.=50°c

s stands for specific heat capacity=?

Mathematically specific heat capacity is given by;

$\rm Q= MC \triangle t \\\\ C = \frac{Q}{M\triangle t} \\\\ C = \frac{3000}{0.465 \triangle 50} \\\\ C =129.0 J/Kg^0C \\\\ C= 0.129 J/g^0C$

Hence the specific heat of the substance will be 0.129 J/g°C.

To learn more about the specific heat capacity refer to the link brainly.com/question/2530523

5 0

1 year ago

A 700 g can of beans is dropped from a shelf that is 1.5 m high. What is the gravitational potential energy of this can? Round y

OLga [1]

<span>To find the gravitational potential energy of an object, we can use this equation: GPE = mgh m is the mass of the object in kg g = 9.80 m/s^2 h is the height of the object in meters GPE = mgh GPE = (0.700 kg) (9.80 m/s^2) (1.5 m) GPE = 10.3 J The gravitational potential energy of this can is 10.3 J</span>

4 0

3 years ago

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