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kati45 [8]
2 years ago
5

If a person driving 130 miles per hour and hit a brick wall ? Does the brick wall hit (push) the car back ? Which Newton laws wo

uld explain this accident and why ?
Physics
2 answers:
belka [17]2 years ago
8 0

An action of push or pull on an object is force. The two bodies interact with each other. Newton's third law of motion states that every action has equal and opposite reaction. This means that if an object A exerts F amount of force on another object B, object B would also exert the same amount of force on A. when a person driving at speed 130 miles per hour hits wall, the wall would also apply an equal force on car and that would cause the damage to it.  

Anuta_ua [19.1K]2 years ago
5 0

Answer#1  

Newton's third law of movement expresses that each activity has equivalent and inverse response. This implies if an item x applies F measure of power on another article y, object y would likewise apply a similar measure of power on x. at the point when an individual driving at speed 130 miles for each hour hits divider, the divider would likewise apply an equivalent power on vehicle and that would make the harm it.  

Answer #2  

Newton's third law of movement will be connected in this situation  

this law expresses that  

each activity has a response meet in sufficiency yet  opposite in indirection.  

This implies if an item state x applies F measure of power on another article y, object y would likewise apply a similar measure of power on x

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svp [43]

Answer:

45

Explanation:

7 0
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

b

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

\frac{m(a+b)^2}{3} \omega_3^2  + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0

\frac{1.53(0.6+0.6)^2}{3} \omega_3^2  + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0

0.7344 \omega_3^2 = 2.128

\omega _3 = \sqrt{\frac{2.128}{0.7344} }

\omega _3 =1.70 \ rad/s

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

7 0
2 years ago
Write a sentence to describe how an image is formed on retina<br>when looking at distant objects.​
beks73 [17]

Answer:

An image is formed on the retina with light rays converging most at the cornea and upon entering and exiting the lens. Rays from the top and bottom of the object are traced and produce an inverted real image on the retina. The distance to the object is drawn smaller than scale

7 0
2 years ago
5.0 kg, with a bullet of mass 0.1 kg. The target was mounted on
kari74 [83]

Answer:

306 m/s

Explanation:

Law of conservation of momentum

m1v1 + m2v2 = (m1+m2)vf

m1 is the bullet's mass so it is 0.1 kg

v1 is what we're trying to solve

m2 is the target's mass so it is 5.0 kg

v2 is the targets velocity, and since it was stationary, its velocity is zero

vf is the velocity after the target is struck by the bullet, so it is 6.0 m/s

plugging in, we get

(0.1 kg)(v1) + (5.0 kg)(0 m/s) = (0.1 kg + 5.0 kg)(6.0 m/s)

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3 years ago
The diagram shows a spherical wave.
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the answer to this question is y

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3 years ago
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