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3 years ago

In which situation is the net force on the object equal to zero?

Physics

1 answer:

Finger [1]3 years ago

7 0

Answer:

A.) a bicycle moving at a constant speed on a straight, level road.

Explanation:

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

$F = ma$

Where;

F represents force measured in Newton.

m represents the mass of an object measured in kilograms.

a represents acceleration measured in meter per seconds square.

In Physics, when the net force acting on a body or an object is equal to zero (0); the body or object is said to be in a state of equilibrium because it is not accelerating. This ultimately implies that, an object whose net force is equal to zero is either moving with a constant speed or the object is static i.e not moving at all.

In this scenario, the situation in which the net force on the object is equal to zero would be a bicycle moving at a constant speed on a straight, level road because it is at equilibrium and as such its resultant force is equal to zero.

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What are radio isotopes??

RideAnS [48]

Answer:

An isotope is one of two or more species of atoms of a chemical element with the same atomic number and position in the periodic table and nearly identical chemical behavior but with different atomic masses and physical properties. In medicine, for example, cobalt-60 is extensively employed as a radiation source to arrest the development of cancer. Other radioactive isotopes are used as tracers for diagnostic purposes as well as in research on metabolic processes.

Hope this helps! Please let me know if you need more help, or if you think my answer is incorrect. Brainliest would be MUCH appreciated. Have a wonderful day!

7 0

2 years ago

Electron guns are used in X-ray tubes. The electrons are accelerated through a relatively large voltage and directed onto a meta

Paul [167]

Answer:

(a) Number of electrons per second strike the target is 3.125 x 10¹⁵.

(b) Charge that strikes the target in 0.750 s is 3.75 x 10⁻⁴ C .

Explanation:

(a) Let n be the number of electrons per second strike the target. We know that current (I) is given by the relation :

I = n x e

Here e is charge of electron.

Substitute 0.500 x 10⁻³ A for I and 1.6 x 10⁻¹⁹ C for e in the above equation.

0.500 x 10⁻³ = n x 1.6 x 10⁻¹⁹

n = 3.125 x 10¹⁵

(b) Let q be the charge that strikes the target. We know that :

q = I x t

Here I is current and t is time.

Substitute 0.500 x 10⁻³ A for I and 0.750 s for t in the above equation,

q = 0.500 x 10⁻³ x 0.750

q = 3.75 x 10⁻⁴ C

3 0

4 years ago

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 13.0 m/s when the

My name is Ann [436]

Answer:

2.82 s

Explanation:

The ball will be subject to the acceleration of gravity which can be considered constant. Therefore we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

Y0 is the starting position, 2.3 m in this case.

Vy0 is the starting speed, 13 m/s.

a will be the acceleration of gravity, -9.81 m/s^2, negative because it points down.

Y(t) = 2.3 + 13 * t - 1/2 * 9.81 * t^2

It will reach the ground when Y(t) = 0

0 = 2.3 + 13 * t - 1/2 * 9.81 * t^2

-4.9 * t^2 + 13 * t + 2.3 = 0

Solving this equation electronically gives two results:

t1 = 2.82 s

t2 = -0.17 s

We disregard the negative solution. The ball spends 2.82 seconds in the air.

7 0

3 years ago

Plzz helpppp meee!!!!

spin [16.1K]

A=f/m
A=900/425
A=2.18

To determine acceleration you divide the force by the mass.

4 0

3 years ago

Read 2 more answers

(1 point) Find a linearly independent set of vectors that spans the same subspace of R3R3 as that spanned by the vectors ⎡⎣⎢3−1−

Natalka [10]

Answer:

Explanation:

$A=\left[\begin{array}{ccc}3&-9&-3\\-1&2&0\\-2&3&-1\end{array}\right] \\\\R_2\rightarrow 3R_2+R_1,R_3\rightarrow 3R_3+2R_1\\\\=\left[\begin{array}{ccc}3&-9&-3\\0&-3&-3\\0&-9&-9\end{array}\right] \\\\R_3\rightarrow 3R_3-9R_2\\\\=\left[\begin{array}{ccc}3&-9&-3\\0&-3&-3\\0&0&0\end{array}\right]$

This is the row echelon form of A. This means that only two of the vectors in our set are linearly independent. In other words, the first two vectors alone will span the same subspace of $R^4$ as all three vectors.

Therefore, the linearly independent spanning set for the subspace is

$\left[\begin{array}{ccc}3\\-1\\-2\end{array}\right] \left[\begin{array}{ccc}-9\\2\\3\end{array}\right] \left[\begin{array}{ccc}3\\0\\-1\end{array}\right]$

5 0

3 years ago