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aleksklad [387]
3 years ago
7

How do positive charges in an electric field move?

Physics
1 answer:
astraxan [27]3 years ago
7 0

Answer:

DOG

Explanation:

jk jk left to right

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an object is moving with a speed of 35 m/s and has a kinetic energy of 1500 J, what is the mass of the object?
Elena-2011 [213]
You'd use the equation kinetic energy=mass*0.5*speed^2
So you'd rearrange this to get mass =kinetic energy /0.5 *speed^2
Which is mass= 1500J/0.5*35^2
=2.44897959183673469........kg
5 0
3 years ago
A skateboarder rolls horizontally off the top of a staircase and lands at the bottom. The staircase has a horizontal length of 1
TiliK225 [7]

Answer: 18.8 m/s

Explanation:

3 0
3 years ago
A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When
natita [175]

Answer:

a) \mu_{k} = 0.704, b) R = 0.312\,m

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s

\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}

\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}

\mu_{k} = 0.704

b) The speed of the block is determined by using the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}

v = x\cdot \sqrt{\frac{k}{m} }

v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }

v \approx 9.829\,\frac{m}{s}

The radius of the circular loop is:

\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}

\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}

R = 0.312\,m

5 0
4 years ago
Doc Brown has calculated his Delorean can accelerate at a rate of 2.52 m/s/s. How
GREYUIT [131]

Answer:

304.89m

Explanation:

Given

acceleration a = 2.52m/s²

final speed v = 39.2m/s

initial speed = 0m/s (car accelerates from rest)

Using the equation of motion below to get the distance of Doc brown from Marty;

v² = u²+2as

substitute the given parameters

39.2² = 0²+2(2.52)s

1536.64 = 0+5.04s

divide both sides by 5.04

1536.64/5.04 = 5.04s/5.04

rearrange the equation

5.04s/5.04 = 1536.64/5.04

s = 304.89m

Hence He and Marty must stand at 304.89m to allow the car to accelerate from rest to a speed  of 39.2 m/s?

6 0
4 years ago
How would be the number 54 be correctly produced in ASL
n200080 [17]
You put your 5 fingers up then you put 4 fingers up on the same hand
3 0
3 years ago
Read 2 more answers
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