Answer:
Yes. Towards the center. 8210 N.
Explanation:
Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.
In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.
The net force is equal to 
Note that 95 km/h is equal to 26.3 m/s.
This is the centripetal force and equal to the x-component of the applied force.
As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.
The amount of the friction force should be 
Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.
Answer:
0.2 m/s
Explanation:
given,
mass of astronaut, M = 85 Kg
mass of hammer, m = 1 Kg
velocity of hammer , v =17 m/s
speed of astronaut, v' = ?
initial speed of the astronaut and the hammer be equal to zero = ?
Using conservation of momentum
(M + m) V = M v' + m v
(M + m) x 0 = 85 x v' + 1 x 17
85 v' = -17
v' = -0.2 m/s
negative sign represent the astronaut is moving in opposite direction of hammer.
Hence, the speed of the astronaut is equal to 0.2 m/s
Answer:
E = 1.655 x 10⁷ N/C towards the filament
Explanation:
Electric field due to a line charge is given by the expression
E = ![[tex]\frac{\lambda}{2\pi\times\epsilon_0\times r}](https://tex.z-dn.net/?f=%5Btex%5D%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Ctimes%5Cepsilon_0%5Ctimes%20r%7D)
[/tex]
where λ is linear charge density of line charge , r is distance of given point from line charge and ε₀ is a constant called permittivity and whose value is
8.85 x 10⁻¹².
Putting the given values in the equation given above
E = 
E = 1.655 x 10⁷ N/C
The frequency of rotation of Mars is 0.0000113 Hertz.
<u>Given the following data:</u>
- Period = 1 day and 37 minutes.
To find the frequency of rotation in Hertz:
First of all, we would convert the the value of period in days and minutes to seconds because the period of oscillation of a physical object is measured in seconds.
<u>Conversion:</u>
1 day = 24 hours
24 hours to minutes = 
× 
= 
minutes
1 minute = 60 seconds
1477 minute = X seconds
Cross-multiplying, we have:
× 
X = 88620 seconds
Now, we can find the frequency of rotation of Mars by using the formula:
<em>Frequency </em><em>of rotation</em> = <em>0.0000113 Hertz</em>
Therefore, the frequency of rotation of Mars is 0.0000113 Hertz.
Read more: brainly.com/question/14708169
a) 10 m/s
b) 25 m
Explanation:
a)
The body is moving with a constant acceleration, therefore we can solve the problem by using the following suvat equation:
where
u is the initial velocity
v is the final velocity
a is the acceleration
t is the time
For the body in this problem:
u = 0 (the body starts from rest)
is the acceleration
t = 5 s is the time
So, the final velocity is
b)
In this second part, we want to calculate the distance travelled by the body.
We can do it by using another suvat equation:
where
u is the initial velocity
v is the final velocity
a is the acceleration
s is the distance travelled
Here we have
u = 0 (the body starts from rest)
is the acceleration
v = 10 m/s is the final velocity
Solving for s,
