Which two substances have no fixed shape and no fixed volume?

Mention two advantage of writing very big and small numbers in tge power of ten.

Elenna [48]

Easier to write, easier to read, easier to understand, easier to compare

7 0

3 years ago

One kind of slingshot consists of a pocket that holds a pebble and is whirled on a circle of radius r. The pebble is released fr

Debora [2.8K]

Answer:

θ=142.9°

Explanation:

d=1 *r

angle ϕ= 37.1°

the line connecting pebble and target should be tangent to a circle so

cos(180-ϕ-θ)= $\frac{r}{d}$ = $\frac{1}{1}$

∴ θ=180-ϕ- $cos^{-1} (\frac{1}{1} )$

θ= 180-37.1-0

θ=142.9°

5 0

3 years ago

A cylindrical, 0.500-m rod has a diameter of 0.02 m. The rod is stretched to a length of 0.501 m by a force of 3000 N. What is t

Salsk061 [2.6K]

Answer:

Y = 4.775 x 10⁹ Pa = 4.775 GPa

Explanation:

First, we calculate the stress on the rod:

$stress = \frac{Force}{Area} = \frac{3000\ N}{\pi r^2} \\\\stress = \frac{3000\ N}{\pi (0.01\ m)^2}\\\\stress = 9.55\ x\ 10^6\ Pa = 9.55 MPa\\$

Now, we calculate the strain:

$strain = \frac{Change\ in Length}{Original\ Length}\\\\strain = \frac{0.501\ m - 0.5\ m}{0.5\ m}\\\\strain = 0.002\\$

Now, we will calculate the Young's Modulus (Y):

$Y = \frac{stress}{strain}\\\\Y = \frac{9.55\ x\ 10^6\ Pa}{0.002} \\$

Y = 4.775 x 10⁹ Pa = 4.775 GPa

6 0

3 years ago

A 1500 kg car is moving on a flat, horizontal flat road. If the radius of the curve is 35 m and

Troyanec [42]

The net force on the car is the friction that keeps it on the road, which points toward the center of the circle of the curve. Then by Newton's second law, we have

• net vertical force:

∑ F = N - W = 0

• net horizontal force:

∑ F = Fs = m a

where

N = magnitude of normal force

W = car's weight

Fs = mag. of static friction

m = car's mass

a = v ²/R = mag. of the centripetal acceleration

v = car's speed

R = radius of curve

Now,

• compute the car's weight:

W = m g = (1500 kg) (9.8 m/s²) = 14,700 N

• solve for the mag. of the normal force:

N = 14,700 N

• solve for the mag. of the friction force, using the given friction coefficient:

Fs = 0.5 N = 7350 N

• solve for the (maximum) acceleration:

7350 N = (1500 kg) a → a = 4.9 m/s²

• solve for the (maximum) speed:

4.9 m/s² = v ²/ (35 m) → v ≈ 13 m/s

4 0

3 years ago

In the year 1178, five monks at Canterbury Cathedral in England observed what appeared to be an asteroid colliding with the moon

Contact [7]

Complete Question

In the year 1178, five monks at Canterbury Cathedral in England observed what appeared to be an asteroid colliding with the moon, causing a red glow in and around it. It is hypothesized that this event created the crater Giordano Bruno, which is right on the edge of the area we can usually see from Earth.

How long after the asteroid hit the Moon,which is $3.84 *10^5 km$ away ,would the light first arrive on the earth in seconds