Answer:
incorrect its 987 for exact
Answer:
734.215N
Explanation:
First we calculate the angle that corresponds to a 5% slope using the Tan-1 function
then we use the component that corresponds to the direction parallel to the road, additionally we must multiply by the gravity value to find the weight(g=9.81m/s^2)
Wx=M*g*sen(2.86)=1500kg*9.81*sen(2.86)=734.215N
There is no need for tangential acceleration when moving in a circle at a constant speed.
<h3>What is centripetal acceleration?</h3>
centripetal acceleration refers to the speed at which a body moves through a circle. Due to the fact that velocity is a vector quantity (i.e., it has both a magnitude, the speed, and a direction), when a body travels in a circle, its direction is constantly changing, which causes a change in velocity, which results in an acceleration.
<h3>Which is an example of centripetal acceleration?</h3>
Centripetal acceleration occurs when you spin a ball on a string above your head. A car experiences centripetal acceleration when it is being driven in a circle. Additionally, a satellite in orbit around the Earth experiences centripetal acceleration.
To know more about tangential acceleration :
brainly.com/question/14993737
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Speed of the car given initially
v = 18 m/s
deceleration of the car after applying brakes will be
a = 3.35 m/s^2
Reaction time of the driver = 0.200 s
Now when he see the red light distance covered by the till he start pressing the brakes
Now after applying brakes the distance covered by the car before it stops is given by kinematics equation
here
vi = 18 m/s
vf = 0
a = - 3.35
so now we will have
So total distance after which car will stop is
So car will not stop before the intersection as it is at distance 20 m
Answer:
83%
Explanation:
On the surface, the weight is:
W = GMm / R²
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.
In orbit, the weight is:
w = GMm / (R+h)²
where h is the height of the shuttle above the surface of the Earth.
The ratio is:
w/W = R² / (R+h)²
w/W = (R / (R+h))²
Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:
w/W = (6.4×10⁶ / 7.03×10⁶)²
w/W = 0.83
The shuttle in orbit retains 83% of its weight on Earth.
