All categories

3 years ago

How fast will a runner be going if he accelerated from rest at 3.0 m/s^2 for a distance of 2.5 m?

Physics

1 answer:

KonstantinChe [14]3 years ago

7 0

Answer:

Vf =3.87[m/s]

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f}^{2}=v_{o}^{2} +2*a*x$

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0 (from the rest)

a = acceleration = 3 [m/s²]

x = distance = 2.5 [m]

Now replacing:

$v_{f}^{2}=0+2*3*2.5\\v_{f}^{2} =15\\v_{f}=\sqrt{15} \\v_{f}=3.87[m/s]$

You might be interested in

While getting buff at the gym you lift a bunch of weights applying 1000N of force to lift them from the ground to a height of 2m

slega [8]

A. 2000 J............

8 0

3 years ago

What is one way you can destroy a magnet’s magnetism?

Alekssandra [29.7K]

-- heat the magnet red-hot in a flame
-- drop it on the floor several times, or beat it with a hammer
-- place it in a strong, rapidly alternating external magnetic field

7 0

3 years ago

A spring that has a force constant of 1050 N/m is mounted vertically on the ground. A block of mass 1.95 kg is dropped from rest

aleksandrvk [35]

Answer:

25.2 cm

Explanation:

K = 1050 N/m

m = 1.95 kg

h = 1.75 m

By the conservation of energy, the potential energy of the block is converted into the potential energy stored in the spring

m g h = 1/2 x k x y^2

Where, y be the distance by which the spring is compressed.

1.95 x 9.8 x 1.75 = 1/2 x 1050 x y^2

33.44 = 525 x y^2

y = 0.252 m

y = 25.2 cm

8 0

3 years ago

Number 25 and 26 please!!!

skad [1K]

laborious ... enclosed may help ... cos -angle = cos +plus angle ... sin - anfle = - sine angle

cos is even function, sine is odd ... draw rhe graohs

6 0

4 years ago

A cord is wrapped around the rim of a solid uniform wheel 0.300 m in radius and of mass 8.00 kg. A steady horizontal pull of 34.

disa [49]

Answer:

A. α = 94.4 rad/s

B. a = 28.32 m/s

C. N = 34N

D. α = 94.4 rad/s

a = 28.32 m/s

N = 44.4 N

Explanation:

part A:

using:

∑T = Iα

where T is the torque, I is the moment of inertia and α is the angular momentum.

firt we will find the moment of inertia I as:

I = $\frac{1}{2}MR^2$

Where M is the mass and R is the radius of the wheel, then:

I = $\frac{1}{2}(8 kg)(0.3 m)^2$

I = 0.36 kg*m^2

Replacing on the initial equation and solving for α, we get::

∑T = Iα

Fr = Iα

34 N = 0.36α

α = 94.4 rad/s

part B

we need to use this equation :

a = αr

where a is the aceleration of the cord that has already been pulled off and r is the radius of the wheel, so replacing values, we get:

a = (94.4)(0.3 m)

a = 28.32 m/s

part C

Using the laws of newton, we know that:

N = T

where N is the force that the axle exerts on the wheel part and T is the tension of the cord

so:

N = 34N

part D

The anly answer that change is the answer of the part D, so, aplying laws of newton, it would be:

-Mg + N +T = 0

Then, solving for N, we get:

N = -T+Mg

N = -34 + (8 kg)(9.8)

N = 44.4 N

6 0

3 years ago