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Genrish500 [490]
2 years ago
12

The velocity of a car is 108 km h-1 what is the velocity in units of ms -1

Physics
1 answer:
Anarel [89]2 years ago
3 0

\frac{108 \times 10^3 }{60 \times 60}  \\  \frac{108 \times 10^3 }{3600} \\  = 30 { ms }^{ - 1}
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juin [17]

Answer:

Following are the solution to this question:

Explanation:

Law:

\to \theta= 180^{\circ}- 50^{\circ}- 25^{\circ}

      = 180^{\circ}- 75^{\circ}\\\\= 105^{\circ}

\to \frac{T_{L}}{\sin (90+50)}= \frac{T_{R}}{\sin (25+90)}=\frac{1400}{\sin (105)}

\to T_L=931.65 \ pounds \\\\ \to T_R=1313.59 \ pounds \\\\

4 0
2 years ago
Someone please help !!
masha68 [24]
I’m pretty sure it’s c.... hope it helps and hope it’s right.
7 0
2 years ago
A proton in a high-energy accelerator is given a kinetic energy of 50.0 GeV. Determine (a) the momentum and (b) the speed of the
koban [17]

(a) The momentum of the proton is determined as 5.17 x 10⁻¹⁸ kgm/s.

(b) The speed of the proton is determined as  3.1 x 10⁹ m/s.

<h3>Momentum of the proton</h3>

The momentum of the proton is calculated as follows;

K.E = ¹/₂mv²

where;

  • m is mass of proton = 1.67 x 10⁻²⁷ kg
  • v is speed of the proton = ?
<h3>Speed of the proton</h3>

v² = 2K.E/m

v² = (2 x 50 x 10⁹ x 1.602 x 10⁻¹⁹ J)/(1.67 x 10⁻²⁷)

v² = 9.6 x 10¹⁸

v = 3.1 x 10⁹ m/s

<h3>Momentum of the proton</h3>

P = mv = (1.67 x10⁻²⁷ x 3.1 x 10⁹) = 5.17 x 10⁻¹⁸ kgm/s

Learn more about momentum here: brainly.com/question/7538238

#SPJ4

4 0
1 year ago
How much work is accomplished when a force of 250 N pushes a box across the floor for a distance of 50 meters?
mote1985 [20]
We use the work formula to solve for the unknown in the problem. The formula for work is expressed as the product of the net force and the distance traveled by the object. We were given both the force and the distance so we can solve work directly.

Work = 250 N x 50 m = 12500 J

Thus, the answer is C.
7 0
2 years ago
How do you know that forces are balanced when static friction acts on an object?
lyudmila [28]
By looking at the acceleration of the object.
In fact, Netwon's second law states that the resultant of the forces acting on an object is equal to the product between the mass m of the object and its acceleration:
\sum F = ma

So, when static friction is acting on the object, if the object is still not moving we know that all the forces are balanced: in fact, since the object is stationary, its acceleration is zero, and so the resultant of the forces (left term in the formula) must be zero as well (i.e. the forces are balanced).
6 0
3 years ago
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