Please help with this question! Thanks in advance...

An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms–1

solniwko [45]

Answer:

Initial velocity of the object, u = 5 m/s

Final velocity of the object, v = 8 m/s

Mass of the object, m = 100 kg

Time take by the object to accelerate, t = 6 s

Initial momentum = mu = 100 — 5 = 500 kg m sˆ’1

Final momentum = mv = 100 — 8 = 800 kg m sˆ’1

Force exerted on the object, F = mv – mu / t

= m (v-u) / t

= 800 – 500

= 300 / 6

= 50 N

Initial momentum of the object is 500 kg m sˆ’1.

Final momentum of the object is 800 kg m sˆ’1.

Force exerted on the object is 50 N.

mark this as brainliest!

6 0

4 years ago

In the picture below identify the type of wave in the top picture. *

Lerok [7]

Answer; transverse wave
Explanation;

In transverse waves the particles vibrate perpendicular to the wave's motion.

4 0

3 years ago

A radar station sends out a 250000 Hz sound wave at a speed of 340 m/s. The sound wave bounces off a weather ballon and returns

Eddi Din [679]

Answers:

a)The balloon is 68 m away of the radar station

b) The direction of the balloon is towards the radar station

Explanation:

We can solve this problem with the Doppler shift equation:

$f'=\frac{V+V_{o}}{V-V_{s}} f$ (1)

Where:

$f=250,000 Hz$ is the actual frequency of the sound wave

$f'=240,000 Hz$ is the "observed" frequency

$V=340 m/s$ is the velocity of sound

$V_{o}=0 m/s$ is the velocity of the observer, which is stationary

$V_{s}$ is the velocity of the source, which is the balloon

Isolating $V_{s}$ :

$V_{s}=\frac{V(f'-f)}{f'}$ (2)

$V_{s}=\frac{340 m/s(240,000 Hz-250,000 Hz)}{240,000 Hz}$ (3)

$V_{s}=-14.16 m/s$ (4) This is the velocity of the balloon, note the negative sign indicates the direction of motion of the balloon: It is moving towards the radar station.

Now that we have the velocity of the balloon (hence its speed, the positive value) and the time ( $t=4.8 s$ ) given as data, we can find the distance:

$d=V_{s}t$ (5)

$d=(14.16 m/s)(4.8 s)$ (6)

Finally:

$d=68 m$ (8) This is the distance of the balloon from the radar station

6 0

3 years ago

Ngan has a weight of 314.5 N on Mars and a weight 833.0 N on Earth. What is Ngan's mass on Earth?

mr_godi [17]

Ngan's mass on earth is 85kg.

Ngan has a weight on Mars = 14.5 N

Ngan’s weight on Earth = 833.0 N

Ngan’s mass on Earth = ?

Fg,earth = mg(earth)

M = Fg,earth / g(earth)

M = 833.0 N / 9.8 m/s2

M = 85 kg

4 0

3 years ago

Read 2 more answers

acontainer is filled whith mercury to alevel of 10m whit water to alevel of 8m and whit oil to alevel of 5m the densities oil ,w

Alborosie

Answer:

1450.4 KN $m^{2}$

Explanation:

Pressure = ρhg

where: ρ is the density of the liquid, h is the height and g the force of gravity.

Total pressure exerted by the liquids at the base = Pressure of oil + Pressure of water + Pressure of mercury

So that,

i. Pressure of oil = ρhg

(ρ = 0.8 g/cm³ = 800 kg/m³)

= 800 x 5 x 9.8

= 39200

Pressure of oil = 39200 N $m^{2}$

ii. Pressure of water = ρhg

(ρ = 1 g/cm³ = 1000 kg/m³)

= 1000 x 8 x 9.8

= 78400

Pressure of water = 78400 N $m^{2}$

ii. Pressure of mercury = ρhg

(ρ = 13.6 g/cm³ = 13600 kg/m³)

= 13600 x 10 x 9.8

= 1332800

Pressure of mercury = 1332800 N $m^{2}$

So that,

Total pressure exerted by the liquids at the base = 39200 + 78400 + 1332800

= 1450400

= 1450.4 KN $m^{2}$

Total pressure exerted by the liquids at the base is 1450.4 KN $m^{2}$ .

8 0

3 years ago