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3 years ago

b. Ron bicycles forward with an acceleration of 2.1 m/s2. If he is applying a forward force of 195 N, what is his mass?

Physics

1 answer:

cricket20 [7]3 years ago

8 0

Answer:

Explanation:

lThe mass of an object given only it's force and acceleration can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question

f = 195 N

a = 2.1 m/s

We have

$m = \frac{195}{2.1 } = 92.857142... \\$

We have the final answer as

Hope this helps you

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A wedge with an inclination of angle θ rests next to a wall. A block of mass m is sliding down the plane. There is no friction b

Softa [21]

Answer:

The net force on the block F(net) = mgsinθ).

Fw =mg(cosθ)(sinθ)

Explanation:

(a)

Here, m is the mass of the block, n is the normal force, \thetaθ is the wedge angle, and Fw is the force exerted by the wall on the wedge.

Since the block sliding down, the net force on the block is along the plane of the wedge that is equal to horizontal component of weight of the block.

F(net) = mgsinθ

The net force on the block F(net) = mgsinθ).

The direction of motion of the block is along the direction of net force acting on the block. Since there is no frictional force between the wedge and block, the only force acting on the block along the direction of motion is mgsinθ.

(b)

From the free body diagram, the normal force n is equal to mgcosθ .

n=mgcosθ

The horizontal component of normal force on the block is equal to force

Fw=n*sin(θ) that exerted by the wall on the wedge.

Substitute mgcosθ for n in the above equation;

Fw =mg(cosθ)(sinθ)

Since, there is no friction between the wedge and the wall, there is component force acting on the wall to restrict the motion of the wedge on the surface and that force is arises from the horizontal component for normal force on the block.

6 0

3 years ago

Un niño amarra una soga a una piedra, y la hace girar como en la gráfica. la piedra realiza un M.C.U, girando con una rapidez de

inessss [21]

Answer:

$\Delta \theta = 56\,rad$

$\Delta \theta \approx 3208.564^{\circ}$

Explanation:

El ángulo barrido en el intervalo de tiempo dado es (The covered angle in the given time interval is):

$\Delta \theta = \omega \cdot \Delta t$

$\Delta \theta = \left(14\,\frac{rad}{s} \rjght)\cdot (4\,s)$

$\Delta \theta = 56\,rad$

$\Delta \theta \approx 3208.564^{\circ}$

4 0

3 years ago

Can someone please help me with this physics question? I'm desperate!

Lelu [443]

Answer:

a) 2·√10 seconds

b) Linda should be approximately 30.6 meters

c) Jenny's speed at the 100-m mark is approximately 6.325 m/s

Explanation:

The speed with which Linda is running = 8.6 m/s

The point Jenny starts = The 80-m mark

The acceleration of Jenny = 1.0 m/s²

a) The time it takes Jenny to run from the 80-m mark to the 100-m mark, <em>t</em>, is given as follows

Δs = u·t + (1/2)·a·t²

Δs = Distance = 100-m - 80-m = 20-m

u = The initial velocity of Jenny = 0

a = Jenny's acceleration = 1.0 m/s²

∴ 20 = 0×t + (1/2) × 1 × t² = t²/2

20 = t²/2

t = √(20 × 2) = 2·√10

The time it takes Jenny to run from the 80-m mark to the 100-m mark = 2·√10 seconds

b) The distance Linda runs in t = 2·√10 seconds, d = v × t

Given that Linda's velocity, v = 8.6 m/s, we have;

d = 8.0 × 2·√10 = 16·√10

The distance Linda runs in t = 2·√10 seconds = 16·√10 meters ≈ 50.6 meters

Therefore, Linda should be approximately (50.6 - 20) meters = 30.6 meters behind Jenny when Jenny starts running

c) Jenny's speed at the 100 m mark is given as follows;

v = u + a·t

t = 2·√10 seconds, a = 1.0 m/s², u = 0

∴ v = 0×t + 1.0×2·√10 = 2·√10 ≈ 6.325

Jenny's speed at the 100-m mark ≈ 6.325 m/s

3 0

3 years ago

A radio have a wavelength of 0.3m and travels at a speed of 300,000,000 m/s. What is the frequency of this wave?

Ilya [14]

The frequency of the wave is $1\cdot 10^9 Hz$

Explanation:

The frequency, the wavelength and the speed of a wave are related by the following equation:

$c=f \lambda$

where

c is the speed of the wave

f is the frequency

$\lambda$ is the wavelength

For the radio wave in this problem,

$\lambda = 0.3 m$

$c=300,000,000 m/s = 3\cdot 10^8 m/s$

Therefore, the frequency is:

$f=\frac{c}{\lambda}=\frac{3\cdot 10^8}{0.3}=1\cdot 10^9 Hz$

Learn more about waves here:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

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3 years ago

what distance is a book from the floor if the book contains 196 joules f potential energy and has a mass of 5 kg?

AleksAgata [21]

E=mgh. 196=5kg*9.81m/s^2*h. So h=196/(5*9.81)=4m

5 0

3 years ago