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3 years ago

A toy car accelerates from 3m/s to 5m/s in 5sec . what is the acceleration

Physics

1 answer:

san4es73 [151]3 years ago

7 0

Answer:

0.4

Explanation:

Because 3m/s is the initial velocity(u) and 5m/s is the final velocity(v) and time is 5 sec.

So, acceleration = v-u ÷ t

I'm confused

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schepotkina [342]

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3 years ago

Read 2 more answers

an object is producing a sound that has a wavelength in air of 2.69m. If the speed of sound in air is 346m/s, what is the freque

Misha Larkins [42]

Answer:

129.74 Hz

Explanation:

Given:

Wave velocity ( v ) = 346 m / sec

wavelength ( λ ) = 2.69 m

We have to calculate Frequency ( f ) :

We know:

v = λ / t [ f = 1 / t ]

v = λ f

= > f = v / λ

Putting values here we get:

= > f = 346 / 2.69 Hz

= > f = 34600 / 269 Hz

= > f = 129.74 Hz

Hence, frequency of sound is 129.74 Hz.

5 0

2 years ago

A rectangular coil with 50 turns of conducting wire and a total resistance of 10.0 Ω initially lies in the yz-plane at time t =

castortr0y [4]

Answer:

a) 43.20V

b) 2.71W/s

c) 40.25s

d) 7.77Nm

Explanation:

(a) The emf of a rotating coil with N turns is given by:

$emf=NBA\omega sin(\omega t)$

N: turns

B: magnitude of the magnetic field

A: area

w: angular velocity

the emf max is given by:

$emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V$

(b) the maximum rate of change of the magnetic flux is given by:

$\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}$

(c) $emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V$

(d) The torque is given by:

$\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm$

3 0

3 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

3 years ago

A student holds a bike wheel and starts it spinning with an initial angular speed of 9.0 rotations per second. The wheel is subj

KATRIN_1 [288]

Answer:

$\Delta t = 8 s$

Explanation:

As we know that the angular acceleration of the wheel due to friction is constant

so we can use kinematics

$\theta = \omega_i t + \frac{1}{2}\alpha t^2$

so we have

$(65 \times 2\pi) = (2\pi \times 9)(10) + \frac{1}{2}(\alpha)(10^2)$

$130\pi = 180\pi + 50 \alpha$

$\alpha = -\pi rad/s^2$

now time required to completely stop the wheel is given as

$\omega_f = \omega_i + \alpha t$

$0 = (2\pi \times 9) + (-\pi) t$

$t = 18 s$

now time required to stop the wheel is given as

$\Delta t = 18 - 10$

$\Delta t = 8 s$

6 0

2 years ago

A toy car accelerates from 3m/s to 5m/s in 5sec . what is the acceleration​

A toy car accelerates from 3m/s to 5m/s in 5sec . what is the acceleration