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sertanlavr [38]
3 years ago
9

an object is hanging by a string from your rearview mirror. While you are decelerating at a constant rate from 25 m/s to rest in

6.0 s (a) what angle does the string make with the vertical? (b) is it toward th windshiled or away from it
Physics
1 answer:
viktelen [127]3 years ago
5 0
This question has already previously been answered. :)

Here it is: brainly.com/question/2141424
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A 20 g bullet is shot from a 10 kg gun with a velocity of 400 m/s. What
Nady [450]

Answer: -.80m/s

Explanation:

3 0
3 years ago
A wave has a frequency of 35 Hz and a wavelength of 15 meters,<br> what is the speed of the wave?
Mila [183]

Answer:

f lamda = c

Explanation:

525 m/s is the speed

5 0
2 years ago
A cheetah can run at a maximum speed 97.8 km/h and a gazelle can run at a maximum speed of 78.2 km/h. If both animals are runnin
emmainna [20.7K]

(1)

Cheetah speed: v_c = 97.8 km/h=27.2 m/s

Its position at time t is given by

S_c (t)= v_c t (1)

Gazelle speed: v_g = 78.2 km/h=21.7 m/s

the gazelle starts S0=96.8 m ahead, therefore its position at time t is given by

S_g(t)=S_0 +v_g t (2)

The cheetah reaches the gazelle when S_c=S_g. Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle:

v_c t=S_0 + v_g t

(v_c -v_g t)=S_0

t=\frac{S_0}{v_c-v_t}=\frac{96.8 m}{27.2 m/s-21.7 m/s}=17.6 s


(2) To solve the problem, we have to calculate the distance that the two animals can cover in t=7.5 s.

Cheetah: S_c = v_c t =(27.2 m/s)(7.5 s)=204 m

Gazelle: S_g = v_g t =(21.7 m/s)(7.5 s)=162.8 m

So, the gazelle should be ahead of the cheetah of at least

d=S_c -S_g =204 m-162.8 m=41.2 m

4 0
3 years ago
Read 2 more answers
Find the mass and center of mass of the solid E with the given density function ρ. E lies under the plane z = 3 + x + y and abov
makvit [3.9K]

Answer:

The mass of the solid is 16 units.

The center of mass of the solid lies at (0.6875, 0.3542, 2.021)

Work:

Density function: ρ(x, y, z) = 8

x-bounds: [0, 1], y-bounds: [0, x], z-bounds: [0, x+y+3]

The mass M of the solid is given by:

M = ∫∫∫ρ(dV) = ∫∫∫ρ(dx)(dy)(dz) = ∫∫∫8(dx)(dy)(dz)

First integrate with respect to z:

∫∫8z(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8x+8y+24](dx)(dy)

Then integrate with respect to y:

∫[8xy+4y²+24y]dx, evaluate y from 0 to x

= ∫[8x²+4x²+24x]dx

Finally integrate with respect to x:

[8x³/3+4x³/3+12x²], evaluate x from 0 to 1

= 8/3+4/3+12

= 16

The mass of the solid is 16 units.

Now we have to find the center of mass of the solid which requires calculating the center of mass in the x, y, and z dimensions.

The z-coordinate of the center of mass Z is given by:

Z = (1/M)∫∫∫ρz(dV) = (1/16)∫∫∫8z(dx)(dy)(dz)

<em>Calculate the integral then divide the result by 16.</em>

First integrate with respect to z:

∫∫4z²(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[4(x+y+3)²](dx)(dy)

= ∫∫[4x²+24x+8xy+4y²+24y+36](dx)(dy)

Then integrate with respect to y:

∫[4x²y+24xy+4xy²+4y³/3+12y²+36y]dx, evaluate y from 0 to x

= ∫[28x³/3+36x²+36x]dx

Finally integrate with respect to x:

[7x⁴/3+12x³+18x²], evaluate x from 0 to 1

= 7/3+12+18

Z = (7/3+12+18)/16 = <u>2.021</u>

The y-coordinate of the center of mass Y is given by:

Y = (1/M)∫∫∫ρy(dV) = (1/16)∫∫∫8y(dx)(dy)(dz)

<em>Calculate the integral then divide the result by 16.</em>

First integrate with respect to z:

∫∫8yz(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8xy+8y²+24y](dx)(dy)

Then integrate with respect to y:

∫[4xy²+8y³/3+12y²]dx, evaluate y from 0 to x

= ∫[20x³/3+12x²]dx

Finally integrate with respect to x:

[5x⁴/3+4x³], evaluate x from 0 to 1

= 5/3+4

Y = (5/3+4)/16 = <u>0.3542</u>

<u />

The x-coordinate of the center of mass X is given by:

X = (1/M)∫∫∫ρx(dV) = (1/16)∫∫∫8x(dx)(dy)(dz)

<em>Calculate the integral then divide the result by 16.</em>

First integrate with respect to z:

∫∫8xz(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8x²+8xy+24x](dx)(dy)

Then integrate with respect to y:

∫[8x²y+4xy²+24xy]dx, evaluate y from 0 to x

= ∫[12x³+24x²]dx

Finally integrate with respect to x:

[3x⁴+8x³], evaluate x from 0 to 1

= 3+8 = 11

X = 11/16 = <u>0.6875</u>

<u />

The center of mass of the solid lies at (0.6875, 0.3542, 2.021)

4 0
2 years ago
A key falls from a bridge that is 45 m above the water. the key falls straight down and lands in a model boat traveling at a con
erastova [34]

Let the key is free falling, therefore from equation of motion

h = ut +\frac{1}{2}gt^2..

Take initial velocity, u=0, so

h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2.

h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2 \\\ t =\sqrt{\frac{2h}{g} }

As velocity moves with constant velocity of 3.5 m/s, therefore we can use formula

d= v \times t

From above substituting t,

d = v \times \sqrt{\frac{2h}{g} }.

Now substituting all the given values and g = 9.8 m/s^2, we get

d = 3.5 \ m/s \times \sqrt{\frac{2 \times 45 m}{9.8 m/s^2} } = 10.60 m.

Thus, the distance the boat was from the point of impact when the key was released is 10.60 m.

7 0
3 years ago
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