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3 years ago

7

Do your relationships tend to have this quality?

2 answers:

dedylja [7]3 years ago

5 0

What quality? /////////

allochka39001 [22]3 years ago

3 0

Oh yes, yes ! Each and every one of my many, many relationships has been obviously and indelibly characterized by that exact quality ! Your question fills me with nothing so much as a portentous sense of deja vu ... all over again !

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A very humble bumble bee is flying horizontally due North at a constant speed of 3.11 m/s. At the current location of the bumble

Reil [10]

To solve this problem we will apply the concepts of the Magnetic Force. This expression will be expressed in both the vector and the scalar ways. Through this second we can directly use the presented values and replace them to obtain the value of the magnitude. Mathematically this can be described as,

$\vec{F_B} = q(\vec{V}\times \vec{B})$

$F_B = q|v||B| sin\theta$

Here,

q = Charge

v = Velocity

B = Magnetic field

$\theta = \text{Angle between } \vec{B} \text{ and } \vec{V}$

Our values are given as,

$\theta = 35.7\°$

$q = 22.5*10^{-9}C$

$B = 1.05*10^{-5}T$

$v = 3.11m/s$

Replacing,

$F_B = (22.5*10^{-9}C)(3.11 \times 1.05*10^{-5}) sin(35.1\°)$

$F_B = 4.224*10^{-13}N$

Therefore the size of the magnetic force acting on the bumble bee is $4.22*10^{-13}N$

3 0

3 years ago

Our galaxy, the ______________, contains more than 200 billion stars.

erica [24]

Answer: Milky Way

Explanation:

4 0

3 years ago

What is stopping distance of a body at 20 m/s is decelerated at 2m/s² to rest?<br>

AlladinOne [14]

Answer:

s = 100 m

Explanation:

v² = u² + 2as

s = (v² - u²) / 2a

s = (0² - 20²) / 2(-2)

s = 100 m

5 0

3 years ago

Determine the final temperature of a gold nugget (mass = 376 g) that starts at 398 K and loses 4.85 kJ of heat to a snowbank whe

m_a_m_a [10]

Answer: Final temperature of a gold nugget will be 297 K

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$q=m\times c\times (T_{final}-T_{initial})$

q = heat released = -4.85 kJ = -4850 J

m = mass of metal = 376 g

$T_{final}$ = final temperature of metal = ?

$T_{initial}$ = initial temperature of metal = 398 K

c = specific heat of metal =

$-4850=376\times 0.128\times (T_{final}-398)$

$(T_{final}-398)=-101$

$(T_{final}=297K$

Thus the final temperature of a gold nugget will be 297 K

6 0

3 years ago

What centripetal force is needed to keep a 7kg mass moving in a circle of 4 meters radius at 15m/s

liq [111]

Answer:

We conclude that the centripetal force needed to keep a 7kg mass moving in a circle of 4 meters radius at 15m/s is 393.75 N.

Explanation:

Given

Mass m = 7 kg

Radius r = 4 m

Velocity v = 15m/s

To determine

We need to determine the centripetal force needed to keep a 7kg mass moving in a circle of 4 meters radius at 15m/s.

We know a centripetal force acts on a body to keep it moving along a curved path.

We can determine the centripetal force using the formula

$F_c=\frac{mv^2}{r}$

where

$m$ is the mass

$v$ is the velocity

$r$ is the radius

$F_c$ is the centripetal force

substitute m = 7, r = 4, and v = 15 in the formula

$F_c=\frac{mv^2}{r}$

$F_c=\frac{7\left(15\right)^2}{4}$

$F_c=\frac{1575}{4}$

$F_c=393.75$ N

Therefore, we conclude that the centripetal force needed to keep a 7kg mass moving in a circle of 4 meters radius at 15m/s is 393.75 N.

7 0

3 years ago