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3 years ago

A vector is In English

Physics

1 answer:

zzz [600]3 years ago

3 0

A vector is a quantity or phenomenon that has two independent properties: magnitude and direction.

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How does our body get heat ? Explain in short . For class 5 plz bro i need help

Murljashka [212]

Answer:

Thermogenesis:Your body's muscles, organs, and brain produce heat in a variety of ways

Explanation:

For example, muscles can produce heat by shivering. Hormonal thermogenesis: Your thyroid gland releases hormones to increase your metabolism. This increases the energy your body creates and the amount of heat it produces

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2 years ago

How does the mass of an object affect the outcome when an unbalanced force acts on it?

Verdich [7]

Answer:

The magnitude of acceleration is reduced.

Explanation:

Force is defined as push or pull

The force is said to be balance force if the force are equal in size but opposite in direction. ie the object does not move or move with constant speed.

The force are to be unbalanced force if the force cause change in motion. ie the object has force greater than zero and has acceleration.

According to Newton second law of motion , acceleration depends on force acting on the object and mass of object.

F=ma

a= $\frac{F}{m}$

When unbalanced force act on the mass of object it reduces magnitude of acceleration without changing the direction.

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3 years ago

a lawn mower is pushed with a force of 50n. if the angle between the handle of the mower and the ground is 30°. why doesn't the

Mademuasel [1]

Answer:

IT doesnt push down because the lawnmower has wheels and the ground is hard

Explanation:

Branliest

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3 years ago

Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at

dedylja [7]

Answer:

$50.91 \mu C$

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for $q_{1}$ and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by $q_{1}$ on q. Then we can know the magnitude of the force exerted by $q_{2}$ about q, finally this will allow us to know the magnitude of $q_{2}$

$q_{1}$ exerts a force on q in +y direction, and $q_{2}$ exerts a force on q in -y direction.

$F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\$

The net force on q is:

$F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}$

Rewriting for $q_{2}$ :

$q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C$

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3 years ago

How do I do number 8, plz respond quick, I have a big unit test on this and I’m rocking a 65 science average rn. Thank you for t

xxTIMURxx [149]

Answer:

3 is the correct answer for number 8

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3 years ago

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