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.
When you squish the spring, you put some energy into it, and after the cord
burns and they go boing in opposite directions, that energy that you stored
in the spring is what gives the blocks their kinetic energy.
But linear momentum still has to be conserved. It was zero while they were
tied together and nothing was moving, so it has to be zero after they both
take off.
Momentum = (mass) x (velocity)
After the launch, the 5.5-kg moves to the right at 6.8 m/s,
so its momentum is
(5.5 x 6.8) = 37.4 kg-m/s to the right.
In order for the total momentum to be zero, the other block has to
carry the same amount of momentum in the opposite direction.
M x V = (6 x speed) = 37.4 kg-m/s to the left.
Divide each side by 6 : Speed = 37.4 / 6 =<em> 6.2333... m/s left</em>
(That number is (6 and 7/30) m/s .)
Answer:
E
Explanation:
Using Coulomb's law equation
Force of the charge = k qQ /d²
and E = F/ q
substitute for F
E = ( K Qq/ d² ) / q
q cancel q
E = KQ / d²
so twice the distance of the from the point charge will lead to the E ( electric field ) decrease by a 4 = E/4. E is inversely proportional to d²
Answer:
The correct option is option (B)
Explanation:
Given that,
The equation below show the total amount of water in gallons, y that has flowed through two different types of showers heads after x minute.
Type A: y =1.25 x
Type B: y= 2.50 x
So,
The amount of water per minute is 
For type A
y =1.25 x
For type B
y =2.50 x
The amount of water flowed through Type B is twice that flowed trough type A per minute.
Answer:
What are you asking me? Tex{t} x1|2?3 squ{R] oo|t?
Explanation:
