All categories

3 years ago

Starting from rest, how far will a brick fall if it is going 15 m/s and accelerates at a rate of 9.8 m/s2?

Physics

1 answer:

masha68 [24]3 years ago

7 0

Answer:

11.48 m

Explanation:

A brick starts from rest and gains a speed of 15 m/s and accelerates at 9.8 m/s^2

u = 0

v= 15

a= 9.8

s= ?

V^2 = U^2 + 2as

15^2 = 0^2 + 2 × 9.8×s

225= 19.6s

s= 225/19.6

s = 11.48m

Hence the brick will fall 11.48 m

You might be interested in

An air conditioner has a power rating of 3000 W and is used daily. Its electrical

Anarel [89]

Answer:

5.714 hours / day

Explanation:

<u>Calculate the hours used in that week </u>

120000 / 3000 = 120 / 3 = 40 hours a week

<u>Calculate the amount it is used in one day</u>

40 / 7 = 5.71428571 hours or 5.714 hours/day

5 0

1 year ago

Read 2 more answers

A negative charge of - 8.0 x 10^-6 C exerts an attractive force of 12 N on a second charge that is

Umnica [9.8K]

Answer:

<h2>Magnitude of the second charge is $-4.17*10^{-7}C$ </h2>

Explanation:

According to columbs law;

F = $kq1q2/r^{2}$

F is the attractive or repulsive force between the charges = 12N

q1 and q2 are the charges

let q1 = - 8.0 x 10^-6 C

q2=?

r is the distance between the charges = 0.050m

k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²

On substituting the given values

12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²

Cross multiplying

$0.03=9*10^{9}* -8.0*10^{-6} q2\\0.03 = -72*10^{3} q2\\q2 = \frac{0.03}{ -72*10^{3}} \\q2 = -4.17*10^{-7}C$

6 0

3 years ago

Read 2 more answers

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

6 0

2 years ago

Justin Bieber is thrown horizontally at 10.0 m/s from the top of a cliff 122.5 m high.

sveticcg [70]

===> Distance fallen from rest in free fall =

   (1/2) (acceleration) (time²)

   (122.5 m) = (1/2) (9.8 m/s²) (time²)

Divide each side by (4.9 m/s²): (122.5 m / 4.9 m/s²) = time²

   (122.5/4.9) s² = time²

Take the square root of each side: 5.0 seconds

===> (Accelerating at 9.8 m/s², he will be dropping at
   (9.8 m/s²) x (5.0 s) = 49 m/s
when he goes 'splat'. We'll need this number for the last part.)

===> With no air resistance, the horizontal component of velocity
doesn't change.

Horizontal distance = (10 m/s) x (5.0 s) = 50 meters .

===> Impact velocity = (10 m/s horizontally) + (49 m/s vertically)

   = √(10² + 49²) = 50.01 m/s arctan(10/49)

   =    50.01 m/s at 11.5° from straight down,
      away from the base of the cliff.

7 0

2 years ago

Read 2 more answers

Which of the following processes is not regarded as contributing to the creation of greenhouse gases? deforestation creating ele

DiKsa [7]

Creating electricity from wind is not regarded as a process contributing to the creation of greenhouse gases. Meanwhile, the processes such as deforestation, the creation of electricity from coal <span>and the use of fertilizers </span><span>are greatly contributing to the making of greenhouse gases.</span>

7 0

2 years ago

Read 2 more answers