A 70 kg box is pulled by a 400 N force at an angle of 30° to the horizontal. The force of kinetic friction is 75 N.

Un auto si muove lungo una strada rettilinea

Valentin [98]

Answer:

Un'auto si muove lungo un percorso rettilineo con velocità variabile come mostrato in figura. Quando l'auto è in possesso di A, la sua velocità è 10 ms-1 e quando è in posizione B, la sua velocità è 20 ms-1. Se l'auto impiega 5 secondi per spostarsi da A a B, trova l'accelerazione dell'auto.

Explanation:

8 0

3 years ago

A 2 kg object being pulled across the floor with a speed of 10 m/sec is suddenly

zvonat [6]

Answer:

The frictional force producing this deceleration would have a magnitude of $4\; \rm N$ .

Explanation:

The velocity of this object changed by $\Delta v = (-10\; \rm m\cdot s^{-1})$ in $\Delta t = 5\; \rm s$ . The acceleration of this object would be:

$\begin{aligned}a &= \frac{\Delta v}{\Delta t} \\ &= \frac{-10\; \rm m\cdot s^{-1}}{5\; \rm s} = -2\; \rm m\cdot s^{-2}\end{aligned}$ .

Let $m$ denote the mass of this object. By Newton's Second Law of Motion, the net force on this object would be:

$\begin{aligned}F &= m \, a \\ &= 2\; \rm kg \times (-2\; \rm m\cdot s^{-2}) \\ &= -4\; \rm N\end{aligned}$ .

( $1\; {\rm kg \cdot m \cdot s^{-2} = 1\; {\rm N}$ .)

If the floor is level, friction would be the only unbalanced force on this object. Thus, the magnitude of the frictional force on this object would also be $4\; {\rm N}$ , same as the magnitude of the net force on this object.

5 0

3 years ago

If one wave of 10 m amplitude meets constructively with another of amplitude 20

steposvetlana [31]

Answer:

B. 30m

Explanation:

This is because when a wave interferes constructively with another wave, they will add since constructive relates to joining or building up.

So, 10m + 20m equals a 30m wave.

3 0

4 years ago

Why on earth the sky is in blue colour but it is black on the moon?

Damm [24]

This is beacuse it is caused by light scattering of light in the atmosphere. When the moon rises blue light is caused and scattered also caused by earth's atmosphere and it is black on the moon as there is no lunar atmosphere , so no light scattering.

7 0

4 years ago

An electron that has a velocity with x component 1.6 × 106 m/s and y component 2.4 × 106 m/s moves through a uniform magnetic fi

Sergio039 [100]

Answer:

(a) 5.056 x 10^-14 N

(b) 5.056 x 10^-14 N

Explanation:

X component of velocity of electron is 1.6 × 10^6 m/s

Y component of velocity of electron is 2.4 × 10^6 m/s

X component of magnetic field is 0.025 T

Y component of magnetic field is -0.16 T

charge on electron, q = - 1.6 x 10^-19 C

Write the velocity and magnetic field in the vector forms.

$\overrightarrow{v}=1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j}$

$\overrightarrow{B}=0.025\widehat{i}-0.16\widehat{j}$

The force on the charge particle when it is moving in the magnetic field is given by

$\overrightarrow{F}=q\left ( \overrightarrow{v}\times \overrightarrow{B} \right )$

(a) Force on electron is given by

$\overrightarrow{F}=-1.6\times 10^{-19}\left ( 1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j} \right )\times \left ( 0.025\widehat{i}-0.16\widehat{j} \right )$

$\overrightarrow{F}=5.056\times 10^{-14}\widehat{k}$

Magnitude of force is 5.056 x 10^-14 N.

(b) Force on a proton is given by

$\overrightarrow{F}=1.6\times 10^{-19}\left ( 1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j} \right )\times \left ( 0.025\widehat{i}-0.16\widehat{j} \right )$

$\overrightarrow{F}=-5.056\times 10^{-14}\widehat{k}$

Magnitude of force is 5.056 x 10^-14 N.

Thus, the magnitude of force remains same but the direction of force is opposite to each other.

Explanation:

4 0

3 years ago