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3 years ago

A cheetah can accelerate from rest at the rate of 4m /s

1 answer:

6 0

Acceleration of cheetah (a) = 4m/s²
time = 10s
initial velocity(u) = 0
final velocity = v
distance travelled = s

v = u +at = 0 + 10×4 = 40m/s
s = (v²-u²)/2a = 40²/(2×4) = 1600/8 = 200m

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Which of the following statements about matter is true? Mass and matter are always the same. Matter is made up of atoms and has

Zepler [3.9K]

The correct answer is Matter is made up of atoms and has mass.

7 0

3 years ago

A wooden disk of mass m and radius r has a string of negligible mass is wrapped around it. If the disk is allowed to fall and th

Tju [1.3M]

Answer:

$a = \frac{2}{3}g$

$T = \frac{mg}{3}$

Explanation:

As the disc is unrolling from the thread then at any moment of the time

We have force equation as

$mg - T = ma$

also by torque equation we can say

$TR = I\alpha$

$TR = \frac{1}{2}mR^2(\frac{a}{R})$

$T = \frac{1}{2}ma$

Now we have

$mg - \frac{1}{2}ma = ma$

$mg = \frac{3}{2}ma$

$a = \frac{2}{3}g$

Also from above equation the tension force in the string is

$T = \frac{1}{2}ma$

$T = \frac{mg}{3}$

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4 years ago

A spring has a spring constant of 120 newtons per meter. Calculate the elastic potential energy stored in the spring when it is

Black_prince [1.1K]

The elastic potential of a spring can be calculated using the formula
E = (1/2) k x²
where are given that
k = 120 N/m
the stretched length is
x = 0.02 m

The elastic potential is
E = (1/2) (120 N/m) (0.02 m)²
E = 0.024 N-m<span />

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3 years ago

The magnetic field due to a utility wire is 0.10 mT when you are at a distance of 10 meters from it. What current (in Amperes) f

ale4655 [162]

Answer:

I = 5000 A

Explanation:

We will use Ampere's Law to calculate the current:

$B = \frac{\mu I}{2\pi r}\\\\$

where,

B = Magnetic Field Strength = 0.1 mT = 1 x 10⁻⁴ T

μ = Permeability of Free Space = 4π x 10⁻⁷ N/A²

I = Current = ?

r = radius = 10 m

Therefore,

$1\ x\ 10^{-4}\ T = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(I)}{2\pi(10\ m)}\\\\I = \frac{(1\ x\ 10^{-4}\ T)(2\pi (10\ m))}{4\pi\ x\ 10^{-7}\ N/A^2}$

4 0

3 years ago

An object initially at rest experiences an acceleration of 0.281 m/s2 to the South for a time of 5.44 seconds. It then increases

andre [41]

Answer:

12.0 meters

Explanation:

Given:

v₀ = 0 m/s

a₁ = 0.281 m/s²

t₁ = 5.44 s

a₂ = 1.43 m/s²

t₂ = 2.42 s

Find: x

First, find the velocity reached at the end of the first acceleration.

v = at + v₀

v = (0.281 m/s²) (5.44 s) + 0 m/s

v = 1.53 m/s

Next, find the position reached at the end of the first acceleration.

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²

x = 4.16 m

Finally, find the position reached at the end of the second acceleration.

x = x₀ + v₀ t + ½ at²

x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²

x = 12.0 m

5 0

3 years ago