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3 years ago

When the motion of one or both of the particles is at an angle to the line of impact, the impact is said to be ________

Physics

1 answer:

Nana76 [90]3 years ago

7 0

Answer: Oblique impact

Explanation:

When the motion of one or both of the particles is at an angle to the line of impact, the impact is said to be oblique impact.

On the other hand, when the directions of motion of the two colliding particles are moving along a line of impact, then it's refered to as central impact.

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What happens in the process of gravitational condensation?

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Answer:

An object decreases in size due to the collision of materials. An object increases in size due to the addition of materials. Gas particles are formed from solar nebula materials.

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3 years ago

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what is the main difference between a substance going through a physical change and one going through a chemical ?

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Answer: Physical changes only change the appearance of a substance, not its chemical composition. Chemical changes cause a substance to change into an entirely substance with a new chemical formula. Chemical changes are also known as chemical reactions.

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2 years ago

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Physics, calculating net force. Please work it out for me

Margarita [4]

Answer:

F = 2985.125 N

Explanation:

Given that,

The radius of curvature of the roller coaster, r = 8 m

Speed of Micheal, v = 17 m/s

Mass of body, m = 65 kg We need to find the net force acting on Micheal. Net force act the bottom of the circle is given by :

$F=\dfrac{mv^2}{r}+mg\\\\F=m(\dfrac{v^2}{r}+g)\\\\=65(\dfrac{17^2}{8}+9.8)\\\\=2985.125\ N$

So, the net force is 2985.125 N.

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3 years ago

A 15.0-kg child descends a slide 2.40 m high and reaches the bottom with a speed of 1.10 m/s .

pickupchik [31]

The thermal energy that is generated due to friction is 344J.

<h3>What is the thermal energy?</h3>

Now we know that the total mechanical energy in the system is constant. The loss in energy is given by the loss in energy.

Thus, the kinetic energy is given as;

KE = 0.5 * mv^2 =0.5 * 15.0-kg * (1.10 m/s)^2 = 9.1 J

PE = mgh = 15.0-kg * 9.8 m/s^2 * 2.40 m = 352.8 J

The thermal energy is; 352.8 J - 9.1 J = 344J

Learn more about thermal energy due to friction:brainly.com/question/7207509

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2 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$