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2 years ago

An interstate highway loop meets the primary route at both ends. the first digit of its route number is always

1 answer:

3 0

The first number indicates that the highway is an East - West or South -North highway. East - West always have an even number and increasing from East going to West while odd numbers are designated to South - North highways with the same incremental pattern.

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A pitcher throws a 0.15 kg baseball so that it crosses home plate horizontally with a speed of 10 m/s. It is hit straight back a

Maru [420]

Answer:

-5.1 kg m/s

Explanation:

Impulse is the change in momentum.

Change in momentum= final momentum - initial momentum=m $v_{2}$ +m $v_{1}$

Plugging in the values= -0.15*24 - (0.15*10) (The motion towards the pitcher is negative as the initial motion is considered to be positive)

Impulse=-5.1 kg m/s (-ve means that it is the impulse towards the pitcher)

4 0

2 years ago

When a mass of 25 g is attached to a certain spring, it makes 20 complete vibrations in 4.0 s. what is the spring constant of th

earnstyle [38]

Answer: The spring of the spring is 25 N/m.

Explanation:

Mass of the body = 25 g= 0.025 kg (1 kg = 1000 g)

Oscillation is 4 sec = 20

Oscillation in 1 sec = $\frac{20}{4}=5$

Frequency of the vibration of the spring = $5 s^{-1}=5 Hz$

Force constant can be calculated bu using the relation between the frequency and, mass and spring constant 'k'

$Frequency=\frac{1}{2\pi}\times \sqrt{\frac{k}{m}}$

$5 s^{-1}=\frac{1}{2\times 3.14}\times \sqrt{\frac{k}{0.025 kg}}$

$k=24.649 N/m\approx 25 N/m$

The spring of the spring is 25 N/m.

3 0

2 years ago

Read 2 more answers

Astronomers were at first surprised to find complicated molecules in the interstellar medium. They thought ultra-violet light fr

jeka57 [31]

Answer:

The dust present in the clouds.

Explanation:

The complicated composition molecules that can be found in space are generally associated with clouds of dust. The significant amount of dust in these clouds provides protection not only for these molecules, but for any body that makes up or is associated with dust clouds.

It is exactly this dust that protects the molecules against the action of ultraviolet rays.

8 0

2 years ago

A 50-n crate sits on a horizontal floor where the coefficient of static friction between the crate and the floor is 0.50 . A 20-

andreyandreev [35.5K]

The resultant static friction force is equal to 20 N to the left.

Why?

I'm assuming that you forgot to write the question of the exercise, so, I will try to complete it:

"A 50-n crate sits on a horizontal floor where the coefficient of static friction between the crate and the floor is 0.50 . A 20-n force is applied to the crate acting to the right. What is the resulting static friction force acting on the crate?"

So, if we are going to calculate the resulting static friction force, it means that there is no movement, we must remember that the friction coefficient will give us the maximum force before the crate starts to move.

We can calculate the static friction force by using the following formula:

$Fr=F(appliedforce)$

Since the crate is not moving (static), the static friction force acting on the crate will be equal to the applied force.

Calculating we have:

$Fr=F(appliedforce)$

$Fr=20N$

Hence, the static friction force is equal to 20 N to the left (since the applied force is acting to the right)

So,

$FrictionForce=AppliedForce$

Since the static friction force is equal to the applied force, the crate does not start to move.

Have a nice day!

8 0

2 years ago

The mass of the Sun is 2multiply1030 kg, and the mass of the Earth is 6multiply1024 kg. The distance from the Sun to the Earth i

spayn [35]

Answer:

a) 3.56 x 10^22 N

b) 3.56 x 10^22 N

Explanation:

Mass of the sun M = 2 x 10^30 kg

mass of the Earth m = 6 x 10^24 kg

Distance between the sun and the Earth R = 1.5 x 10^11 m

From Newton's law,

F = $\frac{GMm}{R^2}$

where F is the gravitational force between the sun and the Earth

G is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2

m is the mass of the Earth

M is the mass of the sun

R is the distance between the sun and the Earth.

Substituting values, we have

F = $\frac{6.67*10^{-11}*2*10^{30}*6*10^{24}}{(1.5*10^{11})^2}$ = 3.56 x 10^22 N

A) The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to 3.56 x 10^22 N

b) The force exerted by the Earth on the Sun = 3.56 x 10^22 N

7 0

2 years ago