All categories

3 years ago

Q1.

Physics

1 answer:

Leno4ka [110]3 years ago

4 0

A) force = mass • acceleration
Convert: 1.870 KN -> 1870 N
1870N = mass • 1.83 m/s^2
———- —————-
1.83 m/s^2. 1.83 m/s^2
-> approx. 1,021.9 grams :D

b) Vi = Vf - at
0 m/s = Vf - 1.83 m/s^2 • 16s
0m/s = Vf - 29.28
+29.28 + 29.28
Vf = 29.28 m/s :D

c) x = Vi + 1/2 • at^2
A to B
T= 15s
Vi = 16m / s
A = 1.83m/s^2
x= 16 + 1/2 • 1.83(15)^2)
x = approx. 221. 9 meters

C to D
T = 9s
Vi= 30m/s
x = 30 + 1/2 1.83(9)^2
x= approx. 104.1 meters

Therefore, when the car is moving at a CONSTANT RATE, it’s traveling farther than when slowing down cause it travel 221.9 meters while slowing down, it’s going only 104.1 meters.

You might be interested in

A war wolf is a device used during the middle ages to assault fortifications with large rocks. A simple trebuchet is constructed

Katarina [22]

Answer:v=41.23 m/s

Explanation:

Given

mass of heavy object $m_1=52 kg$

distance of $m_1$ from the axle $r_1=14 cm$

mass of rock $m_2=123 gm$

Length of rod $=4.1 m$

distance of $m_2$ from axle $r_2=4.1-0.14=3.96 m$

Net torque acting is

$T_{net}=m_1gr_1-m_2gr_2$

$T_{net}=52\times 0.14\times g-0.123\times 3.96\times g$

$T_{net}=6.793\times 9.8$

$T_{net}=66.57 N-m$

Work done by $T_{net}$ is converted to rock kinetic Energy

thus

$T_{net}\times \theta =\frac{mv^2}{2}$

Where $\theta =angle\ turned =\frac{\pi }{2}$

$v= velocity\ at\ launch$

$66.57\times \frac{\pi }{2}=\frac{0.123\times v^2}{2}$

$v^2=66.57\times \pi$

$v=\sqrt{1700.511}$

$v=41.23 m/s$

3 0

4 years ago

This is the smallest unit of matter that cannot be physiologically broken down.

yuradex [85]

An atom is the smallest unit of matter. It can't be broken. It consist of dense nuleus surrounded by negatively charged electrons.

4 0

3 years ago

What traction of the radioisotope<br>remains in the body after one day?

r-ruslan [8.4K]

The fraction of radioisotope left after 1 day is $(\frac{1}{2})^{\frac{1}{\tau}}$ , with the half-life expressed in days

Explanation:

The question is incomplete: however, we can still answer as follows.

The mass of a radioactive sample after a time t is given by the equation:

$m(t)=m_0 (\frac{1}{2})^{\frac{t}{\tau}}$

where:

$m_0$ is the mass of the radioactive sample at t = 0

$\tau$ is the half-life of the sample

This means that the mass of the sample halves after one half-life.

We can rewrite the equation as

$\frac{m(t)}{m_0}=(\frac{1}{2})^{\frac{t}{\tau}}$

And the term on the left represents the fraction of the radioisotope left after a certain time t.

Therefore, after t = 1 days, the fraction of radioisotope left in the body is

$\frac{m(1)}{m_0}=(\frac{1}{2})^{\frac{1}{\tau}}$

where the half-life $\tau$ must be expressed in days in order to match the units.

Learn more about radioactive decay:

brainly.com/question/4207569

brainly.com/question/1695370

#LearnwithBrainly

5 0

3 years ago

A 1750kg bumpercar moving at 1.50m/s to the right collides elastically with a 1450kg car going to the left at 1.10m/s. The 1750k

damaskus [11]

1984.08 kg that’s the answer

6 0

3 years ago

a missile is moving 1810 m/s at a 20.0 degree angle. it needs to hit a target 19,500 m away in a 32.0 degree direction in 9.20 s

Ket [755]

Answer:

112 m/s², 79.1°

Explanation:

In the x direction, given:

x₀ = 0 m

x = 19,500 cos 32.0° m

v₀ = 1810 cos 20.0° m/s

t = 9.20 s

Find: a

x = x₀ + v₀ t + ½ at²

19,500 cos 32.0° = 0 + (1810 cos 20.0°) (9.20) + ½ a (9.20)²

a = 21.01 m/s²

In the y direction, given:

y₀ = 0 m

y = 19,500 sin 32.0° m

v₀ = 1810 sin 20.0° m/s

t = 9.20 s

Find: a

y = y₀ + v₀ t + ½ at²

19,500 sin 32.0° = 0 + (1810 sin 20.0°) (9.20) + ½ a (9.20)²

a = 109.6 m/s²

The magnitude of the acceleration is:

a² = ax² + ay²

a² = (21.01)² + (109.6)²

a = 112 m/s²

And the direction is:

θ = atan(ay / ax)

θ = atan(109.6 / 21.01)

θ = 79.1°

5 0

3 years ago