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nata0808 [166]
2 years ago
6

In what direction does a large mass and small mass object move when affected by unbalanced forces?

Physics
1 answer:
zlopas [31]2 years ago
6 0

Answer: An object's motion can be changed by unbalanced forces. Balanced forces do not change the motion of an object.

The motion of an object acted on by a force depends partly on the strength of the push or pull. The stronger the push or pull, the faster the object will move. For example, the father in the picture below is pushing his daughter in a swing. When he uses more force to push, the swing moves higher.

Explanation:

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A scuba diver measures an increase in pressure of around 10^5 Pa upon descending by 10 m, what is the change in force per square
IceJOKER [234]

Answer:

Explanation:

Given that, .

Pressure around scuba is

P = 10^5 Pa

1 Pa = 1 N/m²

Then

P = 10^5 N/m²

Descending height

h = 10m

Change in force per unit square centimetre

We know that,

Pressure = Force / Area

Then,

Force / Area is the required question we are finding

Then,

Force / Area = 10^5 N / m²

So, let convert the m² to cm²

100cm = 1m

(100cm)² = (1m)²

10⁴cm² = 1m²

Then,

Force / Area = 10^5 N/m² × 1m² / 10⁴cm²

Force / Area = 10 N/cm²

So, the force per unit square centimeters is 10.

7 0
3 years ago
A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 20.1 N; when it is completely i
Paha777 [63]

Answer:

A) V = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}

B) d = 4181.49 kg/m^{3} = 4.18 g/cm^{3}

Explanation:

A) Using the Archimedes' force we can find the weight of water displaced:

W_{d} = W_{a} - W_{w}

Where:

W_{a}: is the weight of the block in the air = 20.1 N

W_{w}: is the weight of the block in the water = 15.3 N

W_{d} = W_{a} - W_{w} = 20.1 N - 15.3 N = 4.8 N

Now, the mass of the water displaced is:

m = \frac{W_{d}}{g} = \frac{4.8 N}{9.81 m/s^{2}} = 0.49 kg

The volume of the block can be found using the mass of water displaced and the density of the water:

V = \frac{m}{d} = \frac{0.49 kg}{997 kg/m^{3}} = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}

B) The density of the block can be found as follows:

d = \frac{W_{a}}{g*V} = \frac{20.1 N}{9.81 m/s^{2}*4.92 \cdot 10^{-4} m^{3}} = 4181.49 kg/m^{3} = 4.18 g/cm^{3}

I hope it helps you!            

6 0
4 years ago
How many times a minute does your heart usually beat?
ycow [4]

Answer:

60-100

Explanation:

A normal resting heart rate for adults ranges from 60 to 100 beats per minute. Generally, a lower heart rate at rest implies more efficient heart function and better cardiovascular fitness.

HOPE THIS HELPS!!! HAVE A GREAT DAY!!!

7 0
3 years ago
Read 2 more answers
A mass spectrometer was used in the discovery of the electron. In the velocity selector, the electric and magnetic fields are se
Mama L [17]

Answer:

Explanation:

Radius of dee, r = 8 mm = 0.008 m

Electric field, e = 400 V/m

Magnetic field, B = 4.7 x 10^-4 T

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

(a) Let v is the speed of electrons.

v = \frac{Bqr}{m}

v = \frac{4.7\times 10^{-4}\times 1.6\times 10^{-19}\times 0.008}{9.1 \times 10^{-31}}

v = 661098.9 = 661099 m/s

(b)

\frac{e}{m}=\frac{1.6 \times 10^{-19}}{9.1\times 10^{-31}}

e / m = 1.76 x 10^14 C / kg

(c) Let K be the kinetic energy

K = 0.5 x mv²

K = 0.5 x 9.1 x 10^-31 x 661099 x 661099

K = 1.99 x 10^-19 J

K = 1.24 eV

So, the potential difference is

V = 1.24 V

(d) if the acceleration voltage is doubled

V = 2 x 1.24 = 2.48 V

So, Kinetic energy

K = 2.48 eV

K = 2.48 x 1.6 x 10^-19 = 3.968 x 10^-19 J

Let v is the speed

K = 0.5 x mv²

3.968 x 10^-19 = 0.5 x 9.1 x 10^-31 x v²

v = 933856.5 m/s

Let the new radius is r.

r=\frac{mv}{Bq}

r=\frac{9.1\times 10^{-31}\times 933856.5}{4.7\times 10^{-4}\times 1.6\times 10^{-19}}

r = 0.0113 m = 1.13 cm

7 0
3 years ago
i throw a tennis ball straight down from the roof of a building 400 meters tall.id it leaves my hand at 10 m/s, how fast will it
Sidana [21]

consider the motion of the tennis ball in downward direction

Y = vertical displacement = 400 m

a = acceleration = acceleration due to gravity = 9.8 m/s²

v₀ = initial velocity of the ball at the top of building = 10 m/s

v = final velocity of the ball when it hits the ground = ?

using the kinematics equation

v² = v²₀ + 2 a Y

inserting the values

v² = 10² + 2 (9.8) (400)

v = 89.11 m/s

6 0
3 years ago
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