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2 years ago

What is an object that can hold things together or lift an object?

Physics

2 answers:

Firdavs [7]2 years ago

4 0

Answer:

Hi myself Shrushtee

Explanation:

Inclined plane = 2. Something that can hold things together or lift an object. Wedge = 3.

Please mark me as brainleist

Mariulka [41]2 years ago

4 0

Answer:

Inclined plane = 2. Something that can hold things together or lift an object. Wedge = 3.

Explanation:

Screw is simple machine which is used to hold to objects together

mark brainliest

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If a river current is 8.0 m/s, and a boat is traveling 10.0 m/s upstream, what is the boat’s speed relative to the riverbank?

Norma-Jean [14]

If the boat is i travling at 10 m/s and the river is 8.0 m/s the boats speed is 18.0 m/s

3 0

3 years ago

Read 2 more answers

A total resistance of 3.03 Ω is to be produced by connecting an unknown resistance to a 12.18 Ω resistance. (a) What must be the

insens350 [35]

Answer:

(a) 4.0334Ω

(b)parallel

Explanation:

for resistors connected in parallel;

$\frac{1}{R_{eq} } =\frac{1}{R1}+\frac{1}{R2}$

Req =3.03Ω , R1 =12.18Ω

$\frac{1}{3.03 } =\frac{1}{12.18}+\frac{1}{R2}$

$\frac{1}{R2}=\frac{1}{3.03 }-\frac{1}{12.18}$

$\frac{1}{R2}=0.2479$

R2=1/0.2479

R2=4.0334Ω

(b)parallel connection is suitable for the desired total resistance. series connection can not be used to achieve a lower resistance as the equation for series connection is.

Req = R1+R2

3 0

3 years ago

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 300 turns. The inner solenoid is 25.0 cm long

joja [24]

Answer:

$5.65487\times 10^{-8}\ Wb$

$1.17\times 10^{-5}\ H$

-0.020475 V

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

$N_1$ = Number of turns of coil = 25

$N_2$ = Number of turns of coil 2 = 300

$\frac{di_2}{dt}$ = Rate of current increased = $1.75\times 10^3\ A/s$

d = Diameter = 2 cm

r = Radius = $\frac{d}{2}=\frac{2}{2}=1\ cm$

A = Area = $\pi r^2$

Magnetic field in the solenoid is given by

$B=\mu_0\frac{N_2}{l}I\\\Rightarrow B=4\pi\times 10^{-7}\frac{300}{0.25}\times 0.12\\\Rightarrow B=0.00018\ T$

Magnetic flux is given by

$\phi=BA\\\Rightarrow \phi=0.00018\times \pi\times 0.01^2\\\Rightarrow \phi=5.65487\times 10^{-8}\ Wb$

The average magnetic flux through each turn of the inner solenoid is $5.65487\times 10^{-8}\ Wb$

Mutual inductance is given by

$L=\frac{N_1\phi}{i_1}\\\Rightarrow L=\frac{25\times 5.65487\times 10^{-8}}{0.12}\\\Rightarrow L=1.17\times 10^{-5}\ H$

The mutual inductance of the two solenoids is $1.17\times 10^{-5}\ H$

Induced emf is given by

$V=-L\frac{di_2}{dt}\\\Rightarrow V=-1.17\times 10^{-5}\times 1.75\times 10^3\\\Rightarrow V=-0.020475\ V$

The emf induced in the outer solenoid by the changing current in the inner solenoid is -0.020475 V

5 0

2 years ago

Describe grafting, what are the<br> three methods?

Tanya [424]

Answer: cleft grafting, inlay grafting, four-flap grafting, and whip grafting.

Explanation:

I hope I helped , Have a great day or night . Xoxoxo.

6 0

2 years ago

Physics Kinematics question

just olya [345]

An interesting problem, and thanks to the precise heading you put for the question.

We will assume zero air resistance.
We further assume that the angle with vertical is t=53.13 degrees, corresponding to sin(t)=0.8, and therefore cos(t)=0.6.

Given:
angle with vertical, t = 53.13 degrees
sin(t)=0.8; cos(t)=0.6;
air-borne time, T = 20 seconds
initial height, y0 = 800 m

Assume g = -9.81 m/s^2

initial velocity, v m/s (to be determined)

Solution:

(i) Determine initial velocity, v.
initial vertical velocity, vy = vsin(t)=0.8v
Using kinematics equation,
S(T)=800+(vy)T+(1/2)aT^2 ....(1)
Where S is height measured from ground.

substitute values in (1): S(20)=800+(0.8v)T+(-9.81)T^2 =>
v=((1/2)9.81(20^2)-800)/(0.8(20))=72.625 m/s for T=20 s

(ii) maximum height attained by the bomb
Differentiate (1) with respect to T, and equate to zero to find maximum
dS/dt=(vy)+aT=0 =>
Tmax=-(vy)/a = -0.8*72.625/(-9.81)= 5.9225 s

Maximum height,
Smax
=S(5.9225)
=800+(0.8*122.625)*(5.9225)+(1/2)(-9.81)(5.9225^2)
= 972.0494 m

(iii) Horizontal distance travelled by the bomb while air-borne
Horizontal velocity = vx = vcos(t) = 0.6v = 43.575 m/s
Horizontal distace travelled, Sx = (vx)T = 43.575*20 = 871.5 m

(iv) Velocity of the bomb when it strikes ground
vertical velocity with respect to time
V(T) =vy+aT...................(2)
Substitute values, vy=58.1 m/s, a=-9.81 m/s^2
V(T) = 58.130 + (-9.81)T =>
V(20)=58.130-(9.81)(20) = -138.1 m/s (vertical velocity at strike)

vx = 43.575 m/s (horizontal at strike)
resultant velocity = sqrt(43.575^2+(-138.1)^2) = 144.812 m/s (magnitude)
in direction theta = atan(43.575,138.1)
= 17.5 degrees with the vertical, downward and forward. (direction)

4 0

3 years ago