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3 years ago

Which substance is a combination of different atoms?

Physics

2 answers:

grigory [225]3 years ago

8 0

Answer:

B. Heterogeneous mixture

Aleks04 [339]3 years ago

8 0

Answer:

It wasn't a heterogeneous mixture, it's compound

Explanation:

I got it wrong on the test when I chose heterogeneous mixture

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A car is moving 5.82M/S when it accelerates at 2.35M/S^2for 3.25S. What is its final velocity?

rodikova [14]

Answer:

Explanation:

vf=vi+at

vi=5.82 m/s

a=2.35 m/s2

t=3.25 s

vf=5.82+2.35*3.25

vf=5.82+7.64

vf=13.46

3 0

3 years ago

A cannonball is fired perfectly horizontally from the top of a 210 m tall cliff. It is fired with an initial velocity of 50 m/s.

pochemuha

Answer:

the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

Explanation:

Given;

height of the cliff, h = 210 m

initial horizontal velocity of the cannonball, Ux = 50 m/s

initial vertical velocity of the cannonball, Uy = 0

The time for the cannonball to reach the ground is calculated as;

$h = u_yt - \frac{1}{2} gt^2\\\\h = 0 - \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 210}{9.8} }\\\\t = 6.55 \ s$

The horizontal distance covered by the cannonball before it hits the ground is calculated as;

$X = U_x \times \ t\\\\X = 50 \times \ 6.55\\\\X = 327.5 \ m$

Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

8 0

3 years ago

Which of the following is a contact force?

erica [24]

Option C Magnetic force

5 0

2 years ago

Read 2 more answers

A projectile is fired vertically upwards and reaches a height of 78.4 m. Find the velocity of projection and the time it takes t

Musya8 [376]

Answer:

1.) U = 39.2 m/s

2.) t = 4s

Explanation: Given that the

height H = 78.4m

The projectile is fired vertically upwards under the acceleration due to gravity g = 9.8 m/s^2

Let's assume that the maximum height = 78.4m. And at maximum height, final velocity V = 0

Velocity of projections can be achieved by using the formula

V^2 = U^2 - 2gH

g will be negative as the object is moving against the gravity

0 = U^2 - 2 × 9.8 × 78.4

U^2 = 1536.64

U = sqrt( 1536.64 )

U = 39.2 m/s

The time it takes to reach its highest point can be calculated by using the formula;

V = U - gt

Where V = 0

Substitute U and t into the formula

0 = 39.2 - 9.8 × t

9.8t = 39.2

t = 39.2/9.8

t = 4 seconds.

7 0

4 years ago

There is a 50 g sample of Ra-229. It has a half-life of 4 minutes.

Sloan [31]

Via half-life equation we have:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h} }$

Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes. By plugging those values in we get:

$A_{final}=50(\frac{1}{2})^\frac{12}{4}=50(\frac{1}{2})^{3}=50(\frac{1}{8})=6.25g$

There is 6.25 grams left of Ra-229 after 12 minutes.

4 0

3 years ago