Ask question

Ask question

All categories

3 years ago

10

Someone help please 1 You are cycling along a flat road. a Are you moving up or down? b What is the force that is pulling you

downwards? c There is a force from the road acting on the bike. Is this force acting upwards or downwards?

1 answer:

9966 [12]3 years ago

7 0

Answer:

$\gamma \: 1a. \: i \: am \: moving \: downwards \: . \\ b. \: gravity \: is \: pulling \: me \: downwards \: \\ c. \: the \: force \: is \: acting \: downwards \: \\ the \: answer \: to \: the \: question \: c \: is \: downwars \: because \: its \: easy \: to \: pull \: the \: bike \: downwards \: but \: its \: hard \: to \: pull \: the \: bike \: upwards \: it \: would \: take \: more \: energy \: \gamma \\ \\$

Explanation:

<h3><em> $\infty \infty \: hope \: it \: helps \: you \: \\ have \: a \: good \: a \: day \: .$ </em></h3>

<em><u>S</u></em><em><u>l</u></em><em><u>i</u></em><em><u>d</u></em><em><u>e</u></em><em><u> </u></em><em><u>m</u></em><em><u>y</u></em><em><u> </u></em><em><u>a</u></em><em><u>n</u></em><em><u>s</u></em><em><u>w</u></em><em><u>e</u></em><em><u>r</u></em><em><u> </u></em><em><u>t</u></em><em><u>o</u></em><em><u> </u></em><em><u>s</u></em><em><u>e</u></em><em><u>e</u></em><em><u> </u></em><em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u> </u></em><em><u>f</u></em><em><u>u</u></em><em><u>l</u></em><em><u>l</u></em><em><u> </u></em><em><u>a</u></em><em><u>n</u></em><em><u>s</u></em><em><u>w</u></em><em><u>e</u></em><em><u>r</u></em><em><u> </u></em><em><u>.</u></em><em><u> </u></em>

You might be interested in

Convert 5.6kg to grams

OLEGan [10]

Answer:

my pleasure

Explanation:

5.6kg = 5600 grams

multiply the mass value by 1000

4 0

3 years ago

Read 2 more answers

What studies the natural world around us

Tanzania [10]

Science studies the natural world. This includes the components of the physical universe around us like atoms, plants, ecosystems, people, societies and galaxies, as well as the natural forces at work on those things.

<h3><u>PLEASE</u><u> MARK</u><u> ME</u><u> BRAINLIEST</u><u>.</u></h3>

6 0

3 years ago

Read 2 more answers

The slower the particles in a substance move,

swat32

Answer: particles move closer together

Explanation: If the motion of particles slows the particles move closer together. This is because the attraction between them pulls them toward each other. Strong attractive forces hold particles close together. As the motion of particles increases, particles move further apart.

3 0

3 years ago

An object of mass 6 kg. is resting on a horizontal surface. A horizontal force

son4ous [18]

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force ( $W$ ), measured in joules, is defined by the following expression:

$W = F\cdot \Delta x$ (1)

Where:

$F$ - Force, measured in newtons.

$\Delta x$ - Distance, measured in meters.

If we know that $F = 15\,N$ and $\Delta x = 100\,m$ , then the work done by the force exerted on the object is:

$W = (15\,N)\cdot (100\,m)$

$W = 1500\,J$

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object ( $a$ ), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

$\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}$ (2)

Where $v_{o}$ is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

$\frac{1}{2}\cdot a \cdot t^{2} = \Delta x-v_{o}\cdot t$

$a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}$

If we know that $\Delta x = 100\,m$ , $v_{o} = 0\,\frac{m}{s}$ and $t = 10\,s$ , then the net acceleration experimented by the object is:

$a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}$

$a = 2\,\frac{m}{s^{2}}$

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

$\Sigma F = F - f = m\cdot a$ (3)

Where:

$F$ - External force exerted on the object, measured in newtons.

$f$ - Kinetic friction force, measured in newtons.

If we know that $F = 15\,N$ , $m = 6\,kg$ and $a = 2\,\frac{m}{s^{2}}$ , the kinetic friction force is:

$f = F-m\cdot a$

$f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)$

$f = 3\,N$

The work done by friction ( $W'$ ), measured in joules, is:

$W' = f\cdot \Delta x$ (4)

$W' = (3\,N) \cdot (100\,m)$

$W' = 300\,J$

And the net work experimented by the object is:

$\Delta W = 1500\,J - 300\,J$

$\Delta W = 1200\,J$

By the Work-Energy Theorem we understand that change in translational kinetic energy ( $\Delta K$ ), measured in joules, is equal to the change in net work. That is:

$\Delta K = \Delta W$ (5)

If we know that $\Delta W = 1200\,J$ , then the change in translational kinetic energy is:

$\Delta K = 1200\,J$

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

7 0

3 years ago

In which electric circuit would the voltmeter read 10 volts ?

ki77a [65]

Given that,

Voltage = 10 volt

Suppose, The three resistance is connected in parallel and each resistance is 12 Ω. find the current in the electric circuit.

We need to calculate the equivalent resistance

Using formula of parallel

$\dfrac{1}{R}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}$

Put the value into the formula

$\dfrac{1}{R}=\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}$

$\dfrac{1}{R}=\dfrac{1}{4}$

$R=4\ \Omega$

We need to calculate the current in the circuit

Using ohm's law

$V=IR$

$I=\dfrac{V}{R}$

Where, V = voltage

R = resistance

Put the value into the formula

$I=\dfrac{10}{4}$

$I=2.5\ A$

Hence, The current in the circuit is 2.5 A

4 0

3 years ago