All categories

2 years ago

Ton is used for measure very big masses.right or wrongscience

Physics

2 answers:

nlexa [21]2 years ago

4 0

Answer:

Explanation:

The correct answer is right

adoni [48]2 years ago

3 0

Answer:

True

Explanation:

Ton is a unit that is used to measure huge masses like cars, trucks and huge boxes

You might be interested in

Two long, straight wires are separated by 0.120 m. The wires carry currents of 11 A in opposite directions, as the drawing indic

rewona [7]

Answer:

The magnitude of the magnetic field is 1.83 x $10^{-5}$ T.

Explanation:

The flow of an electric current in a straight wire induces magnetic field around the wire. When current is flowing through two wires in the same direction, a force of attraction exists between the wires. But if the current flows in opposite directions, the force of repulsion is felt by the wires.

In the given question, the direction of flow of current through the wires is opposite, thus both wires applies the same field on each other. The result to repulsion between them.

The magnetic field (B) between the given wires can be determined by:

B = $\frac{U_{o}I }{2\pi r}$

where: I is the current, r is the distance between the wires and $U_{0}$ is the magnetic field constant.

But, I = 11 A, r = 0.12 m and $U_{0}$ = 4 $\pi$ x $10^{-7}$ Tm/A

So that;

B = $\frac{4\pi *10^{-7}*11 }{2\pi *0.12}$

= 1.8333 x $10^{-5}$

B = 1.83 x $10^{-5}$ T

6 0

2 years ago

If a 990 kg car is on the road and the Ff is 360 n what is the normal force

Veseljchak [2.6K]

Given that a car is in the road, there is only movement in the x-direction. There is no movement in the y-direction.

Looking at the y-direction for the normal force:

F = N - mg
0 = N - mg, (no movement in y-dir.)
N = mg
N = (990)(9.8)
N = 9702 newtons

The normal force exerted on the car by the road is 9702 newtons.

8 0

3 years ago

Read 2 more answers

A small rock is thrown vertically upward with a speed of 17.0m/s from the edge of the roof of a 26.0m tall building. The rock do

Pachacha [2.7K]

Answer:

A) v = 28.3 m/s

B) t = 4.64 s

Explanation:

Assuming no other forces acting on the rock, since the accelerarion due to gravity close to the surface to the Earth can be taken as constant, we can use one of the kinematic equations in order to get first the maximum height (over the roof level) that the ball reaches:

$v_{f}^{2} - v_{o}^{2} = 2* g* \Delta h (1)$

Taking into account that at this point, the speed of the rock is just zero, this means vf=0 in (1), so replacing by the givens and solving for Δh, we get:

$\Delta h = \frac{-v_{o} ^{2}}{2*g} = \frac{-(17.0m/s)^{2} }{2*(-9.8m/s2)} = 14.8 m (2)$

So, we can use now the same equation, taking into account that the initial speed is zero (when it starts falling from the maximum height) and that the total vertical displacement is the distance between the roof level and the ground (26.0 m) plus the maximum height that we have just found in (2) , 14.8m:
Δh = 26.0 m + 14. 8 m = 40.8 m (3)
Replacing now in (1), we can solve for vf, as follows:

$v_{f} =\sqrt{2*g*\Delta h} = \sqrt{2*9.8m/s2*40.8m} = 28.3 m/s (4)$

In order to find the total elapsed from when the rock is thrown until it hits the street, we can divide this time in two parts:
1) Time elapsed from the the rock is thrown, until it reaches to its maximum height, when vf =0
2) Time elapsed from this point until it hits the street, with vo=0.
For the first part, we can simply use the definition of acceleration (g in this case), making vf =0, as follows:

$v_{f} = v_{o} + a*\Delta t = v_{o} - g*\Delta t = 0 (5)$

Replacing by the givens in (5) and solving for Δt, we get:

$\Delta t = \frac{v_{o}}{g} = \frac{17.0m/s}{9.8m/s2} = 1.74 s (6)$

For the second part, since we know the total vertical displacement from (3), and that vo = 0 since it starts to fall, we can use the kinematic equation for displacement, as follows:

$\Delta h = \frac{1}{2} * g * t^{2} (7)$

Replacing by the givens and solving for t in (7), we get:

$t_{fall} =\sqrt{\frac{2*\Delta h}{g}} = \sqrt{\frac{2*40.8m}{9.8m/s2} } = 2.9 s (8)$

So, total time is just the sum of (6) and (8):
t = 2.9 s + 1.74 s = 4.64 s

5 0

2 years ago

Why is Pizza so good?

Allushta [10]

I honestly don’t know-

6 0

3 years ago

Read 2 more answers

An archer pulls the bowstring back for a distance of 0.470 m before releasing the arrow. The bow and string act like a spring wh

never [62]

Answer:

(a) 46.94 J.

(b) 55.95 m/s

Explanation:

(a)

Potential Energy: This is the energy of a body, due to its position. The S.I unit of potential energy is Joules (J).

The formula of potential energy in a stretched spring is

Ep = 1/2ke² .......................... Equation 1

Where Ep = potential energy of the spring, k = Force constant of the spring, e = extension or compression.

Given: k = 425 N/m, e = 0.47 m.

Substitute into equation 1

Ep = 1/2(425×0.47²)

Ep = 46.94 J.

(b)

at the instant When the arrow leaves the bow, the potential energy of the arrow is converted kinetic energy of the bow.

I.e,

Ep = 1/2mv² ............. Equation 2

Where m = mass of the arrow, v = velocity of the arrow.

make v the subject of the equation

v = √(2Ep/m)............. Equation 3

Given: Ep = 46.94 J, m = 0.03 Kg.

Substitute into equation 3

v = √(2×46.96/0.03)

v = √(93.92/0.03)

v = √(3130.67)

v = 55.95 m/s

5 0

2 years ago

Ton is used for measure very big masses.right or wrongscience ​

Ton is used for measure very big masses.right or wrongscience