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3 years ago

Force F=5 N acts to the right

Physics

1 answer:

icang [17]3 years ago

8 0

Answer:

Calculations involving forces

The resultant force is the single force that has the same effect as two or more forces acting together.

Two forces in the same direction

Two forces that act in the same direction produce a resultant force that is larger than either individual force.

You can easily calculate the resultant force of two forces that act in a straight line in the same direction by adding their sizes together.

Example

Two forces, 3 N and 2 N, act to the right. Calculate the resultant force.

Two arrows, one above the other, both pointing to the right, one labelled 2 N and one labelled 3 N. Then an equals sign and then another arrow to the right labelled 5 N.

Two forces acting in the same direction

Resultant force F = 3 N + 2 N = 5 N to the right.

The resultant force is 5 N to the right.

Two forces in opposite directions

Two forces that act in opposite directions produce a resultant force that is smaller than either individual force.

To find the resultant force subtract the magnitude of the smaller force from the magnitude of the larger force.

The direction of the resultant force is in the same direction as the larger force.

Example

A force of 5 N acts to the right, and a force of 3 N act to the left.

Calculate the resultant force.

Two arrows, one above the other, one pointing to the left, labelled 2 N, the other pointing to the right labelled 3 N. Then an equals sign, with an arrow to the right labelled 1 N.

Two forces acting in opposite directions

Resultant force F

Resultant force F = 5 N - 3 N = 2 N to the right.

The resultant force is 2 N to the right.

Explanation:

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Suppose a family replaces ten 60-watt incandescent bulbs with ten 30-watt

mixer [17]

Initially there were 10 bulbs of 60 Watt power

So total power of all bulbs = 60 * 10 = 600 W

now each bulb used for 4 hours daily

so total energy consumed daily

E = p*t

E = 600*4 = 2400Wh

E = 2.4kWH

now we have total power consumed in 1 year

E = 365 * 2.4 = 876kWh

cost of electricity = 10 cents/ kWh

so total cost of energy for one year

P1 = 876*10 = 8760cents = 87.60

Now if all 60 Watt bulbs are replaced by 30 Watt bulbs

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3 years ago

A snail travels 300 cm in 4 minutes.calculate speed of snail in m/s

LekaFEV [45]

Answer:

$\frac{0.0125m}{s}$

Explanation:

In order to solve this question we need to know that $speed = \frac{distance}{time}$ . Then we need to convert 4 minutes into seconds and cm into m. We can do that by multiplying the number of minutes by 60 (because there is 60 seconds in one minute) and dividing the number of cm by 100 (because there is 100 cm in one m). So.......

4min = 4 x 60s = 240s

300cm = 300/100 m = 3m

Now we know that distance = 300m, and that the time = 4min = 240s ⇒

⇒ $speed=\frac{distance}{time} = \frac{3m}{240s} = \frac{0.0125m}{s}$

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3 years ago

A firearms company is testing a new model of rifle by firing a 7.50-g lead bullet into a block of wood having a mass of 17.5 kg.

Anestetic [448]

Answer:

Explanation:

Let the bullets speed be V .

Kinetic energy = 1/2 mV² where m is mass of bullet

This energy is converted into heat Q which raises the temperature of target by Δ T .

Q = mc Δ T , m is mass , c is specific heat and Δ T is rise in temperature .

heat absobed by bullet

= .0075 x 130 x .040

= .039 J

heat absorbed by block of wood

= 17.5 x 1700 x .04

= 1190 J

Total heat absorbed

= 1190.039 J

So kinetic energy = heat absobed

= 1/2 x .0075 x V² = 1190.039

V² = 317343.73

V = 563.33 m /s

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3 years ago

What is the path of object due to gravitational forces

fiasKO [112]

if drppoed then locally path vertical down ... towards centre of earth

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3 years ago

For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par

Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

initial temperature of metals, = $T_m$
initial temperature of water, = $T_i$
specific heat capacity of copper, $C_p$ = 0.385 J/g.K
specific heat capacity of aluminum, $C_A$ = 0.9 J/g.K
both metals have equal mass = m

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

For copper metal, the quantity of heat transferred is calculated as;

$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

The change in heat energy for copper metal;

$\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t$

For aluminum metal, the quantity of heat transferred is calculated as;

$Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w) = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t$

The change in heat energy for aluminum metal ;

$\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t$

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Learn more here:brainly.com/question/15345295

6 0

3 years ago

Force F=5 N acts to the right ​

Force F=5 N acts to the right