A bus manufacturer decides to double a window's area in order to give passengers a

Can someone please help me I don’t understand this please show work plus I got it wrong it’s not C

matrenka [14]

B,
The 25 N and 5 N force are acting in the same direction so we can add them together, but the 10 N force acts in the opposite direction so you subtract it.
25 N + 5 N - 10 N = 20 N

8 0

3 years ago

An arcade ball is thrown with an initial speed of 7.0 m/s and follows the trajectory shown. The ball enter the basket .95 second

dmitriy555 [2]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

$x = 4.70 \ m$ , $y = 0.2803 \ m$

Explanation:

From the question we are told that

The initial velocity of the arcade ball is $u = 7.0 \ m/s$

The time taken by the ball to enter the basket is $t = 0.95 \ s$

Generally the x -component of the initial velocity is

$u_x = 7 * cos (45)$

=> $u_x = 4.95 \ m/s$

Generally the y -component of the initial velocity is

$u_x = 7 * sin(45)$

=> $u_y = 4.95 \ m/s$

Generally the distance x is mathematically represented as

$x = u_x * t$

=> $x = 4.95 * 0.95$

=> $x = 4.70 \ m$

Generally from kinematic equation the distance y is mathematically represented as

$y = u_y * t - \frac{1}{2} * g * t^2$

=> $y = 4.95 * 0.95 - \frac{1}{2} * 9.8 * 0.95^2$

=> $y = 0.2803 \ m$

7 0

3 years ago

A 4-kg block and a 2-kg block can move on the horizontal frictionless surface. The blocks are accelerated by a+12-N force that p

lozanna [386]

Answer:

$F_n = -4 N$

Explanation:

Net external force that exerted on the block is given as

$F_{net} = (m_1 + m_2) a$

here we know that

$F = 12 N$

$m_1 = 4 kg$

$m_2 = 2 kg$

now we have

$12 = (4 + 2) a$

so we have

$a = 2 m/s^2$

now the force exerted by bigger block on smaller block is given as

$F_n = ma$

$F_n = (2 \times 2)$

$F_n = 4 N$

now we know that two blocks will exert equal and opposite force on each other

so here the force exerted by 2 kg block on 4 kg block will be

$F_n = -4 N$

5 0

3 years ago

Two 1 MHz radio antennas emitting in-phase are separated by600

algol [13]

Answer:

1 km

Explanation:

$d_0$ = Gap between antennas = 600 m

$\nu$ = Frequency = 1 MHz

$z_1$ = Distance to receiver = 2 km

c = Speed of light = $3\times 10^8\ m/s$

Wavelength is given by

$\lambda=\dfrac{c}{\nu}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{1\times 10^6}\\\Rightarrow \lambda=300\ m$

Distance to be moved is given by

$D=\dfrac{z_1\lambda_0}{d_0}\\\Rightarrow D=\dfrac{2\times 300}{600}\\\Rightarrow D=1\ km$

The distance to be moved is 1 km north.

6 0

3 years ago

If Q = 16 nC, a = 3.0 m, and b = 4.0 m, what is the magnitude of the electric field at point P?

Lynna [10]

In BPC

tan\theta =a/b = 3/4

\theta = tan^-1(0.75)

\theta = 36.87 deg

BP = sqrt(a^2 + b^2) = sqrt((3)^2 + (4)^2) = 5 m

Eb = k Q/BP^2 = (9 x 10^9) (16 x 10^-9)/5^2 = 5.76 N/C

Ea = k Q/AP^2 = (9 x 10^9) (16 x 10^-9)/4^2 = 9 N/C

Ec = k Q/CP^2 = (9 x 10^9) (16 x 10^-9)/3^2 = 16 N/C

Net electric field along X-direction is given as

Ex = Ea + Eb Cos36.87 = (9) + (5.76) Cos36.87 = 13.6 N/C

Net electric field along X-direction is given as

Ey = Ec + Eb Sin36.87 = (16) + (5.76) Sin36.87 = 19.5 N/C

Net electric field is given as

E = sqrt(Ex^2 + Ey^2) = sqrt((13.6)^2 + (19.5)^2) = 23.8 N/C

8 0

3 years ago