The weight of a rock is 8.5 pounds. Does this statement include

Consider a Cassegrain-focus, reflecting telescope. Images recorded at Cassegrain-focus will be:

Contact [7]

B. Flipped compared to what is in the sky

3 0

3 years ago

A 2-kg cart, traveling on a horizontal air track with a speed of 3m/s, collides with a stationary 4-kg cart. The carts stick tog

ladessa [460]

Answer:

The impulse exerted by one cart on the other has a magnitude of 4 N.s.

Explanation:

Given;

mass of the first cart, m₁ = 2 kg

initial speed of the first car, u₁ = 3 m/s

mass of the second cart, m₂ = 4 kg

initial speed of the second cart, u₂ = 0

Let the final speed of both carts = v, since they stick together after collision.

Apply the principle of conservation of momentum to determine v

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2 x 3 + 0 = v(2 + 4)

6 = 6v

v = 1 m/s

Impulse is given by;

I = ft = mΔv = m(

The impulse exerted by the first cart on the second cart is given;

I = 2 (3 -1 )

I = 4 N.s

The impulse exerted by the second cart on the first cart is given;

I = 4(0-1)

I = - 4 N.s (equal in magnitude but opposite in direction to the impulse exerted by the first).

Therefore, the impulse exerted by one cart on the other has a magnitude of 4 N.s.

8 0

3 years ago

The mass of a radioactive substance follows a continuous exponential decay model, with a decay rate parameter of 1% per day. A s

Bumek [7]

Answer:

2,38kg

Explanation:

Mass in function of time can be found by the formula: $m_{(t)} =m_{0} e^{-kt}$ , where $m_{0}$ is the initial mass, t is the time and k is a constant.

Given that a sample decay 1% per day, that means that after first day you have 99% of mass.

$m_{(1)} =m_{0} e^{-k(1)}$ , but $m_{(1)}=\frac{99m_{0} }{100}$ , so we have $\frac{99m_{0} }{100}=m_{0}e^{-k}$ , then $k=-ln(\frac{99}{100})=0.01$

Now using k found we must to find $m_{(5)}$ .

$m_{(5)}=m_{0}e^{-(0.01)5}=2.5kge^{-0.05} =2.5x0.951=2.38kg$

6 0

3 years ago

The x vector component of a displacement vector has a magnitude of 86.2 m and points along the negative x axis. The y vector com

Flauer [41]

Answer:

(a) Magnitude of Vector = 207.73 m

(b) Direction = 65.48°

Explanation:

(a)

The formula to find out the magnitude of a resultant vector with the help of its x and y components is given as follows:

$Magnitude\ of\ Vector = \sqrt{d_{x}^{2} + d_{y}^{2}} = \sqrt{(86.2\ m)^{2} + (189\ m)^{2}}\\\\$

<u>Magnitude of Vector = 207.73 m</u>

(b)

For the direction of the vector we have the formula:

$Direction = tan^{-1}(\frac{y}{x})\\\\Direction = tan^{-1}(\frac{189\ m}{86.2\ m})\\\\$

<u>Direction = 65.48°</u>

4 0

3 years ago

Which word is not a good way to describe exercise?

Karolina [17]

In my opinion, calling exercise "work" wouldn't be a good way of describing it, I don't think people would exercise if its called work

3 0

3 years ago