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2 years ago

A change in position with respect to a reference point is called?

Physics

1 answer:

kumpel [21]2 years ago

4 0

A change in position with respect to a reference point is called motion

hope it helps...

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during a journey, a car travels at 40 km in 2.5 hours, next 62 km in 3 hours, then took a break for 30 minutes, again travelled

wlad13 [49]

45mph is the answer if you do the math right

6 0

2 years ago

ball of mass 0.4 kg is attached to the end of a light stringand whirled in a vertical circle of radius R = 2.9 m abouta fixed po

Levart [38]

Answer:

$6.046N$

Explanation:

The net force exerted on the mass is the sum of tension force and the external force of gravity.

$F_n_e_t=F_g+F_t$

$F_t$ is the tension force. $F_g=9.8N/kg$ is the force of gravity.

$F_n_e_t=ma_c=mv^2/r\\$

where $r$ is the rope's radius from the fixed point.

From the net force equation above:

$F_t=F_n_e_t-F_g\\=mv^2/r-mg\\=0.4\times(8.5^2/2.9)-0.4\times9.8\\=6.046N$

Hence the tension force is 6.046N

8 0

2 years ago

At a point 1.2 m out from the hinge, 14.0 N force is exerted at an angle of 27 degrees to the moment arm in a plane which is per

geniusboy [140]

Answer:

$\tau = 7.63 Nm$

Explanation:

As we know that moment of force is given as

$\tau = \vec r \times \vec F$

now we have

$\vec r = 1.2 m$

$\vec F = 14 N$

now from above formula we have

$\tau = r F sin\theta$

here we know that

$\theta = 27 degree$

so we have

$\tau = (1.2)(14) sin27$

$\tau = 7.63 Nm$

3 0

2 years ago

Una carga de -10Mc está situada a 20cm delante de otra carga de 5 Mc. Calcular la fuerza electrostática en Newton ejercida por u

tino4ka555 [31]

Answer:

a) force between them is attraction, b) F = 1.125 10⁻² N

Explanation:

In this case the electric force is given by Coulomb's law

F = $k \frac{q_1q_2}{r^2}$

In the exercise they give us the values of the loads

q1 = - 10 mC = -10 10⁻³ C

q2 = 5 mC = 5 10⁻³ C

d = 20 cm = 0.20 m

let's calculate

F = 9 10⁹ $\frac{10 \ 10^{-3} \ 5 \ 10^{-3}}{0.20^2}$

F = 1.125 10⁻² N

To find the direction of the force we use that charges of the same sign repel each other, as in this case there is a positive and a negative charge, the force between them is attraction

7 0

2 years ago

A football is kicked from the ground with a velocity of 38m/s at an angle of 40 degrees and eventually lands at the same height.

Anastasy [175]

Initially, the velocity vector is $\langle 38cos(40^{\circ}),38sin(40^{\circ}) \rangle=\langle 29.110, 24.426 \rangle$ . At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by $4.9(0.2)^2$ , so the velocity is $\langle 29.110, 24.426-0.196 \rangle = \langle 29.110, 24.23 \rangle$ .

Converting back to direction and magnitude, we get $\langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle$

4 0

2 years ago