You push the ball with aforce of 22.8N which induces a -2.3N frictional force. What is the net force while you push?

The four wheels of a car are connected to the car's body by spring assemblies that let the wheels move up and down over bumps an

Marrrta [24]

Answer: $k=55.590 KN/m$

Explanation:

Given

mass of person $\left ( m\right )$ =68 kg

car dips about 1.2 cm

We know

F=kx

Where k=combined spring constant

mg=kx

$k=\frac{mg}{x}$

$k=\frac{68\times 9.81}{1.2\time 10^{-2}}$

$k=55.590 KN/m$

7 0

4 years ago

The graph shows a wave that oscillates with a frequency of 60 Hz. Based on the information given in the diagram, what is the spe

Snezhnost [94]

Answer:

900 cm/s or 9 m/s.

Explanation:

Data obtained from the question include the following:

Length (L) = 30 cm

frequency (f) = 60 Hz

Velocity (v) =.?

Next, we shall determine the wavelength (λ).

This is illustrated below:

Since the wave have 4 node, the wavelength of the wave will be:

λ = 2L/4

Length (L) = 30 cm

wavelength (λ) =.?

λ = 2L/4

λ = 2×30/4

λ = 60/4

λ = 15 cm

Therefore, the wavelength (λ) is 15 cm

Now, we can obtain the speed of the wave as follow:

wavelength (λ) = 15 cm

frequency (f) = 60 Hz

Velocity (v) =.?

v = λf

v = 15 × 60

v = 900 cm/s

Thus, converting 900 cm/s to m/s

We have:

100 cm/s = 1 m/s

900 cm/s = 900/100 = 9 m/s

Therefore, the speed of the wave is 900 cm/s or 9 m/s.

5 0

3 years ago

A block is held at rest against a wall by a force of magnitude F exerted at an angle theta from the horizontal, as shown in the

wel

Answers:

B.) $F cos\theta=F_{n}$

C.) $F sin\theta=F_{g} \pm F_{f}$

Explanation:

The image attached shows the way the force $F$ is acting on the block. Now, if we draw a free body diagram of the situation and write the equations for the Net Force in X and Y, we will have the following:

Net Force in X:

$-F_{n}+F cos\theta=0$ (1)

Where:

$F_{n}$ is the Normal force

$F$ is the magnitude of the force exerted on the block

$\theta$ is the angle

Net Force in Y:

$F sin\theta \pm F_{f}-F_{g}=0$ (2)

Where:

$F_{f}$ is the Friction force (it is expresed with the $\pm$ sign because this force may be up or down, we cannot know because the block is at rest)

$F_{g}$ is the gravity force

Rewrittin (1):

$F cos\theta=F_{n}$ (3) This is according to option B

Rewritting (2):

$F sin\theta=F_{g}\pm F_{f}$ (3) This is according to option C

3 0

3 years ago

Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in

charle [14.2K]

Answer:

94.13 ft/s

Explanation:

<u>Given:</u>

$t$ = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s

$s$ = distance to be moved by the rock long the horizontal = 98 yards

$y$ = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

<u>Assume:</u>

$u$ = magnitude of initial velocity of the rock

$\theta$ = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

$\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\$

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

$\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\$

On dividing equation (1) by (2), we have

$\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ$

Now, putting this value in equation (2), we have

$u\cos 51.34^\circ\times 5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s$

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

3 0

3 years ago

I meed help with these 2 questions plz

ludmilkaskok [199]

-- 30N

-- the sum of all the forces

4 0

3 years ago