v₀ = initial velocity of the mobile = 10 m/s
v = final velocity of the mobile = 20 m/s
a = acceleration of the mobile = 5 m/s²
d = distance traveled during this operation = ?
Using the kinematics equation
v² = v²₀ + 2 a d
inserting the above values in the equation
20² = 10² + 2 (5) d
400 = 100 + 10 d
subtracting 100 both side
400 - 100 = 100 - 100 + 10 d
300 = 10 d
dividing both side by 10
300/10 = 10 d/10
d = 30 m
hence mobile travels 30 m.
Answer: The answer is C.) 25 m/s^2.
Explanation: If you input 5 as s, you would have to use the exponent 2. This means that you have to multiply 5 by 5. 5 x 5= 25.
Edit: Also, because the surface is frictionless, it will make the object go faster too. Nothing can really slow it down unless something blocks it.
Answer:
a.) L = 2.64 kgm^2/s
b.) V = 4.4 m/s
Explanation: Jessica stretches her arms out 0.60 m from the center of her body. This will be considered as radius.
So,
Radius r = 0.6 m
Mass M = 2 kg
Velocity V = 1.1 m/s
Angular momentum L can be expressed as;
L = MVr
Substitute all the parameters into the formula
L = 2 × 1.1 × 0.6 = 1.32kgm^2s^-1
the combined angular momentum of the masses will be 2 × 1.32 = 2.64 kgm^2s-1
b. If she pulls her arms into 0.15 m,
New radius = 0.15 m
Using the same formula again
L = 2( MVr)
2.64 = 2( 2 × V × 0.15 )
1.32 = 0.3 V
V = 1.32/0.3
V = 4.4 m/s
Her new linear speed will be 4.4 m/s
The cart travelled a distance of 14.4 m
Explanation:
The work done by a force when pushing an object is given by:

where:
F is the magnitude of the force
d is the displacement
is the angle between the direction of the force and the displacement
In this problem we have:
W = 157 J is the work done on the cart
F = 10.9 N is the magnitude of the force
, assuming the force is applied parallel to the motion of the cart
Therefore we can solve for d to find the distance travelled by the cart:

Learn more about work:
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